How to replace epoch timestamps in a file with other formats?

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up vote
9
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I have a file that contains epoch dates which I need converting to human-readable. I already know how to do the date conversion, eg:



[server01 ~]$ date -d@1472200700
Fri 26 Aug 09:38:20 BST 2016


..but I'm struggling to figure out how to get sed to walk through the file and convert all the entries.
The file format looks like this:



#1472047795
ll /data/holding/email
#1472047906
cat /etc/rsyslog.conf
#1472048038
ll /data/holding/web









share|improve this question



















  • 1




    For future reference (assuming this is a Bash history file; it looks like one), look to the HISTTIMEFORMAT shell variable to control the format at time of writing.
    – Toby Speight
    Aug 30 '16 at 14:44










  • @Toby the value of HISTTIMEFORMAT is used when displaying (to stdout), but only its status (set to anything even null vs unset) matters when writing HISTFILE.
    – dave_thompson_085
    Sep 3 '16 at 10:43










  • Thanks @dave, I didn't know that (not being a user of history times myself).
    – Toby Speight
    Sep 5 '16 at 9:00










  • date -d is not portable to say Solaris... I'm assuming this is on a system with mostly GNU tools? (GNU AWK / Perl tend to be the more portable methods to deal with date conversions). gawk ' if ($0 ~ /^#[0-9]*$/) print strftime("%c",substr($0,2)); else print ' < file (strftime seems non-portable...)
    – Gert van den Berg
    Jan 13 '17 at 8:07














up vote
9
down vote

favorite
2












I have a file that contains epoch dates which I need converting to human-readable. I already know how to do the date conversion, eg:



[server01 ~]$ date -d@1472200700
Fri 26 Aug 09:38:20 BST 2016


..but I'm struggling to figure out how to get sed to walk through the file and convert all the entries.
The file format looks like this:



#1472047795
ll /data/holding/email
#1472047906
cat /etc/rsyslog.conf
#1472048038
ll /data/holding/web









share|improve this question



















  • 1




    For future reference (assuming this is a Bash history file; it looks like one), look to the HISTTIMEFORMAT shell variable to control the format at time of writing.
    – Toby Speight
    Aug 30 '16 at 14:44










  • @Toby the value of HISTTIMEFORMAT is used when displaying (to stdout), but only its status (set to anything even null vs unset) matters when writing HISTFILE.
    – dave_thompson_085
    Sep 3 '16 at 10:43










  • Thanks @dave, I didn't know that (not being a user of history times myself).
    – Toby Speight
    Sep 5 '16 at 9:00










  • date -d is not portable to say Solaris... I'm assuming this is on a system with mostly GNU tools? (GNU AWK / Perl tend to be the more portable methods to deal with date conversions). gawk ' if ($0 ~ /^#[0-9]*$/) print strftime("%c",substr($0,2)); else print ' < file (strftime seems non-portable...)
    – Gert van den Berg
    Jan 13 '17 at 8:07












up vote
9
down vote

favorite
2









up vote
9
down vote

favorite
2






2





I have a file that contains epoch dates which I need converting to human-readable. I already know how to do the date conversion, eg:



[server01 ~]$ date -d@1472200700
Fri 26 Aug 09:38:20 BST 2016


..but I'm struggling to figure out how to get sed to walk through the file and convert all the entries.
The file format looks like this:



#1472047795
ll /data/holding/email
#1472047906
cat /etc/rsyslog.conf
#1472048038
ll /data/holding/web









share|improve this question















I have a file that contains epoch dates which I need converting to human-readable. I already know how to do the date conversion, eg:



[server01 ~]$ date -d@1472200700
Fri 26 Aug 09:38:20 BST 2016


..but I'm struggling to figure out how to get sed to walk through the file and convert all the entries.
The file format looks like this:



#1472047795
ll /data/holding/email
#1472047906
cat /etc/rsyslog.conf
#1472048038
ll /data/holding/web






text-processing sed date






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share|improve this question













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edited Nov 18 at 10:05









Rui F Ribeiro

38.2k1475123




38.2k1475123










asked Aug 30 '16 at 7:38









machinist

6716




6716







  • 1




    For future reference (assuming this is a Bash history file; it looks like one), look to the HISTTIMEFORMAT shell variable to control the format at time of writing.
    – Toby Speight
    Aug 30 '16 at 14:44










  • @Toby the value of HISTTIMEFORMAT is used when displaying (to stdout), but only its status (set to anything even null vs unset) matters when writing HISTFILE.
    – dave_thompson_085
    Sep 3 '16 at 10:43










  • Thanks @dave, I didn't know that (not being a user of history times myself).
    – Toby Speight
    Sep 5 '16 at 9:00










  • date -d is not portable to say Solaris... I'm assuming this is on a system with mostly GNU tools? (GNU AWK / Perl tend to be the more portable methods to deal with date conversions). gawk ' if ($0 ~ /^#[0-9]*$/) print strftime("%c",substr($0,2)); else print ' < file (strftime seems non-portable...)
    – Gert van den Berg
    Jan 13 '17 at 8:07












  • 1




    For future reference (assuming this is a Bash history file; it looks like one), look to the HISTTIMEFORMAT shell variable to control the format at time of writing.
    – Toby Speight
    Aug 30 '16 at 14:44










  • @Toby the value of HISTTIMEFORMAT is used when displaying (to stdout), but only its status (set to anything even null vs unset) matters when writing HISTFILE.
    – dave_thompson_085
    Sep 3 '16 at 10:43










  • Thanks @dave, I didn't know that (not being a user of history times myself).
    – Toby Speight
    Sep 5 '16 at 9:00










  • date -d is not portable to say Solaris... I'm assuming this is on a system with mostly GNU tools? (GNU AWK / Perl tend to be the more portable methods to deal with date conversions). gawk ' if ($0 ~ /^#[0-9]*$/) print strftime("%c",substr($0,2)); else print ' < file (strftime seems non-portable...)
    – Gert van den Berg
    Jan 13 '17 at 8:07







1




1




For future reference (assuming this is a Bash history file; it looks like one), look to the HISTTIMEFORMAT shell variable to control the format at time of writing.
– Toby Speight
Aug 30 '16 at 14:44




For future reference (assuming this is a Bash history file; it looks like one), look to the HISTTIMEFORMAT shell variable to control the format at time of writing.
– Toby Speight
Aug 30 '16 at 14:44












@Toby the value of HISTTIMEFORMAT is used when displaying (to stdout), but only its status (set to anything even null vs unset) matters when writing HISTFILE.
– dave_thompson_085
Sep 3 '16 at 10:43




@Toby the value of HISTTIMEFORMAT is used when displaying (to stdout), but only its status (set to anything even null vs unset) matters when writing HISTFILE.
– dave_thompson_085
Sep 3 '16 at 10:43












Thanks @dave, I didn't know that (not being a user of history times myself).
– Toby Speight
Sep 5 '16 at 9:00




Thanks @dave, I didn't know that (not being a user of history times myself).
– Toby Speight
Sep 5 '16 at 9:00












date -d is not portable to say Solaris... I'm assuming this is on a system with mostly GNU tools? (GNU AWK / Perl tend to be the more portable methods to deal with date conversions). gawk ' if ($0 ~ /^#[0-9]*$/) print strftime("%c",substr($0,2)); else print ' < file (strftime seems non-portable...)
– Gert van den Berg
Jan 13 '17 at 8:07




date -d is not portable to say Solaris... I'm assuming this is on a system with mostly GNU tools? (GNU AWK / Perl tend to be the more portable methods to deal with date conversions). gawk ' if ($0 ~ /^#[0-9]*$/) print strftime("%c",substr($0,2)); else print ' < file (strftime seems non-portable...)
– Gert van den Berg
Jan 13 '17 at 8:07










6 Answers
6






active

oldest

votes

















up vote
6
down vote



accepted










Assuming consistent file format, with bash you can read the file line by line, test if it's in given format and then do the conversion:



while IFS= read -r i; do [[ $i =~ ^#([0-9]10)$ ]] && 
date -d@"$BASH_REMATCH[1]"; done <file.txt


BASH_REMATCH is an array whose first element is the first captured group in Regex matching, =~, in this case the epoch.




If you want to keep the file structure:



while IFS= read -r i; do if [[ $i =~ ^#([0-9]10)$ ]]; then printf '#%sn' 
"$(date -d@"$BASH_REMATCH[1]")"; else printf '%sn' "$i"; fi; done <file.txt


this will output the modified contents to STDOUT, to save it in a file e.g. out.txt:



while ...; do ...; done >out.txt


Now if you wish, you can replace the original file:



mv out.txt file.txt



Example:



$ cat file.txt
#1472047795
ll /data/holding/email
#1472047906
cat /etc/rsyslog.conf
#1472048038
ll /data/holding/web

$ while IFS= read -r i; do [[ $i =~ ^#([0-9]10)$ ]] && date -d@"$BASH_REMATCH[1]"; done <file.txt
Wed Aug 24 20:09:55 BDT 2016
Wed Aug 24 20:11:46 BDT 2016
Wed Aug 24 20:13:58 BDT 2016

$ while IFS= read -r i; do if [[ $i =~ ^#([0-9]10)$ ]]; then printf '#%sn' "$(date -d@"$BASH_REMATCH[1]")"; else printf '%sn' "$i"; fi; done <file.txt
#Wed Aug 24 20:09:55 BDT 2016
ll /data/holding/email
#Wed Aug 24 20:11:46 BDT 2016
cat /etc/rsyslog.conf
#Wed Aug 24 20:13:58 BDT 2016
ll /data/holding/web





share|improve this answer






















  • Nice....that prints the converted date to screen, now how do I get that command to replace the entries in the file?
    – machinist
    Aug 30 '16 at 8:01










  • @machinist Check my edits..
    – heemayl
    Aug 30 '16 at 8:17






  • 1




    If you are using a recent version of bash, printf can do the conversion itself: printf '#%(%F %H)Tn' "$BASH_REMATCH[1]".
    – chepner
    Aug 30 '16 at 19:54


















up vote
14
down vote













While it's possible with GNU sed with things like:



sed -E 's/^#([0-9]+).*$/date -d @1/e'


That would be terribly inefficient (and is easy to introduce arbitrary command injection vulnerabilities1) as that would mean running one shell and one date command for each #xxxx line, virtually as bad as a shell while read loop. Here, it would be better to use things like perl or gawk, that is text processing utilities that have date conversion capabilities built-in:



perl -MPOSIX -pe 's/^#(d+).*/ctime $1/se'


Or:



gawk '/^#/$0 = strftime("%c", substr($0, 2));1'



1 If we had written ^#([0-9]).* instead of ^#([0-9]).*$ (as I did in an earlier version of this answer), then in multi-byte locales like UTF-8 ones (the norm nowadays), with an input like #1472047795<0x80>;reboot, where that <0x80> is the byte value 0x80 which does not form a valid character, that s command would have ended up running date -d@1472047795<0x80>; reboot for instance. While with the extra $, those lines would not be substituted. An alternative approach would be: s/^#([0-9])/date -d @1 #/e, that is leave the part after the #xxx date as a shell comment






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  • What about using just a single instance of date -f to do all the conversions in a stream-wise manner?
    – Digital Trauma
    Aug 30 '16 at 20:58










  • The perl command seems to add a new line after ctime $1 and I can't find any way to remove it.
    – Alex Harvey
    Jun 23 '17 at 7:47







  • 1




    @Alex. Right. See edit. Adding the s flag makes so that .* also includes the newline on input. You can also use strftime "%c", localtime $1.
    – Stéphane Chazelas
    Jun 23 '17 at 14:02










  • @StéphaneChazelas thanks so much. It's a great answer.
    – Alex Harvey
    Jun 23 '17 at 14:04

















up vote
2
down vote













All the other answers spawn a new date process for every epoch date that needs to be converted. This could potentially add performance overhead if your input is large.



However GNU date has a handy -f option that allows a single process instance of date to continuously read input dates without the need for a new fork. So we can use sed, paste and date in this manner such that each one only gets spawned once (2x for sed) regardless of how large the input is:



$ paste -d 'n' <( sed '2~2d;y/#/@/' epoch.txt | date -f - ) <( sed '1~2d' epoch.txt )
Wed Aug 24 07:09:55 PDT 2016
ll /data/holding/email
Wed Aug 24 07:11:46 PDT 2016
cat /etc/rsyslog.conf
Wed Aug 24 07:13:58 PDT 2016
ll /data/holding/web
$


  • The two sed commands respectively basically delete even and odd lines of the input; the first one also replaces # with @ to give the correct epoch timestamp format.

  • The first sed output is then piped to date -f which does the required date conversion, for every line of input that it receives.

  • These two streams are then interlaced into the single required output using paste. The <( ) constructs are bash process substitutions that effectively trick paste into thinking it is reading from given filenames when it is in fact reading the output piped from the command inside. -d 'n' tells paste to separate odd and even output lines with a newline. You could change (or remove) this if for example you want the timestamp on the same line as the other text.

Note that there are several GNUisms and Bashisms in this command. This is not Posix-compliant and should not be expected to be portable outside of the GNU/Linux world. For example date -f does something else on OSXes BSD date variant.






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  • date -d (from the question) is also non-portable... (On FreeBSD it will try to mess with DST settings, on Solaris it will give an error...) The question does not specify an OS though...
    – Gert van den Berg
    Jan 13 '17 at 8:14










  • @GertvandenBerg yes, this is addressed in the last paragraph of this answer.
    – Digital Trauma
    Jan 13 '17 at 15:56










  • I mean that the asker's sample also has portability issues... (They should probably have tagged an OS...)
    – Gert van den Berg
    Jan 14 '17 at 16:06

















up vote
1
down vote













Assuming the date format you have in your post is what you want, the following regex should fit your needs.



sed -E 's/#(1[0-9]9)(.*)/echo 1 $(date -d @1)/e' log.file


Be mindful of the fact this will only replace one epoch per line.






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  • I'm getting the following error with that command: sed: -e expression #1, char 48: invalid reference 3 on 's' command's RHS
    – machinist
    Aug 30 '16 at 7:53







  • 1




    My mistake, edited the post.
    – Hatclock
    Aug 30 '16 at 8:25

















up vote
0
down vote













using sed :



sed -r 's/#([0-9]*)/echo $(date -d @1)/eg' test.txt


output :



ر أغس 24 16:09:55 EET 2016
ll /data/holding/email
ر أغس 24 16:11:46 EET 2016
cat /etc/rsyslog.conf
ر أغس 24 16:13:58 EET 2016
ll /data/holding/web


as my locale language is Arabic :)






share|improve this answer





























    up vote
    0
    down vote













    My solution how to do that in a pipeline



    cat test.txt | sed 's/^/echo "/; s/([0-9]10)/`date -d @1`/; s/$/"/' | bash





    share|improve this answer




















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      6 Answers
      6






      active

      oldest

      votes








      6 Answers
      6






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      6
      down vote



      accepted










      Assuming consistent file format, with bash you can read the file line by line, test if it's in given format and then do the conversion:



      while IFS= read -r i; do [[ $i =~ ^#([0-9]10)$ ]] && 
      date -d@"$BASH_REMATCH[1]"; done <file.txt


      BASH_REMATCH is an array whose first element is the first captured group in Regex matching, =~, in this case the epoch.




      If you want to keep the file structure:



      while IFS= read -r i; do if [[ $i =~ ^#([0-9]10)$ ]]; then printf '#%sn' 
      "$(date -d@"$BASH_REMATCH[1]")"; else printf '%sn' "$i"; fi; done <file.txt


      this will output the modified contents to STDOUT, to save it in a file e.g. out.txt:



      while ...; do ...; done >out.txt


      Now if you wish, you can replace the original file:



      mv out.txt file.txt



      Example:



      $ cat file.txt
      #1472047795
      ll /data/holding/email
      #1472047906
      cat /etc/rsyslog.conf
      #1472048038
      ll /data/holding/web

      $ while IFS= read -r i; do [[ $i =~ ^#([0-9]10)$ ]] && date -d@"$BASH_REMATCH[1]"; done <file.txt
      Wed Aug 24 20:09:55 BDT 2016
      Wed Aug 24 20:11:46 BDT 2016
      Wed Aug 24 20:13:58 BDT 2016

      $ while IFS= read -r i; do if [[ $i =~ ^#([0-9]10)$ ]]; then printf '#%sn' "$(date -d@"$BASH_REMATCH[1]")"; else printf '%sn' "$i"; fi; done <file.txt
      #Wed Aug 24 20:09:55 BDT 2016
      ll /data/holding/email
      #Wed Aug 24 20:11:46 BDT 2016
      cat /etc/rsyslog.conf
      #Wed Aug 24 20:13:58 BDT 2016
      ll /data/holding/web





      share|improve this answer






















      • Nice....that prints the converted date to screen, now how do I get that command to replace the entries in the file?
        – machinist
        Aug 30 '16 at 8:01










      • @machinist Check my edits..
        – heemayl
        Aug 30 '16 at 8:17






      • 1




        If you are using a recent version of bash, printf can do the conversion itself: printf '#%(%F %H)Tn' "$BASH_REMATCH[1]".
        – chepner
        Aug 30 '16 at 19:54















      up vote
      6
      down vote



      accepted










      Assuming consistent file format, with bash you can read the file line by line, test if it's in given format and then do the conversion:



      while IFS= read -r i; do [[ $i =~ ^#([0-9]10)$ ]] && 
      date -d@"$BASH_REMATCH[1]"; done <file.txt


      BASH_REMATCH is an array whose first element is the first captured group in Regex matching, =~, in this case the epoch.




      If you want to keep the file structure:



      while IFS= read -r i; do if [[ $i =~ ^#([0-9]10)$ ]]; then printf '#%sn' 
      "$(date -d@"$BASH_REMATCH[1]")"; else printf '%sn' "$i"; fi; done <file.txt


      this will output the modified contents to STDOUT, to save it in a file e.g. out.txt:



      while ...; do ...; done >out.txt


      Now if you wish, you can replace the original file:



      mv out.txt file.txt



      Example:



      $ cat file.txt
      #1472047795
      ll /data/holding/email
      #1472047906
      cat /etc/rsyslog.conf
      #1472048038
      ll /data/holding/web

      $ while IFS= read -r i; do [[ $i =~ ^#([0-9]10)$ ]] && date -d@"$BASH_REMATCH[1]"; done <file.txt
      Wed Aug 24 20:09:55 BDT 2016
      Wed Aug 24 20:11:46 BDT 2016
      Wed Aug 24 20:13:58 BDT 2016

      $ while IFS= read -r i; do if [[ $i =~ ^#([0-9]10)$ ]]; then printf '#%sn' "$(date -d@"$BASH_REMATCH[1]")"; else printf '%sn' "$i"; fi; done <file.txt
      #Wed Aug 24 20:09:55 BDT 2016
      ll /data/holding/email
      #Wed Aug 24 20:11:46 BDT 2016
      cat /etc/rsyslog.conf
      #Wed Aug 24 20:13:58 BDT 2016
      ll /data/holding/web





      share|improve this answer






















      • Nice....that prints the converted date to screen, now how do I get that command to replace the entries in the file?
        – machinist
        Aug 30 '16 at 8:01










      • @machinist Check my edits..
        – heemayl
        Aug 30 '16 at 8:17






      • 1




        If you are using a recent version of bash, printf can do the conversion itself: printf '#%(%F %H)Tn' "$BASH_REMATCH[1]".
        – chepner
        Aug 30 '16 at 19:54













      up vote
      6
      down vote



      accepted







      up vote
      6
      down vote



      accepted






      Assuming consistent file format, with bash you can read the file line by line, test if it's in given format and then do the conversion:



      while IFS= read -r i; do [[ $i =~ ^#([0-9]10)$ ]] && 
      date -d@"$BASH_REMATCH[1]"; done <file.txt


      BASH_REMATCH is an array whose first element is the first captured group in Regex matching, =~, in this case the epoch.




      If you want to keep the file structure:



      while IFS= read -r i; do if [[ $i =~ ^#([0-9]10)$ ]]; then printf '#%sn' 
      "$(date -d@"$BASH_REMATCH[1]")"; else printf '%sn' "$i"; fi; done <file.txt


      this will output the modified contents to STDOUT, to save it in a file e.g. out.txt:



      while ...; do ...; done >out.txt


      Now if you wish, you can replace the original file:



      mv out.txt file.txt



      Example:



      $ cat file.txt
      #1472047795
      ll /data/holding/email
      #1472047906
      cat /etc/rsyslog.conf
      #1472048038
      ll /data/holding/web

      $ while IFS= read -r i; do [[ $i =~ ^#([0-9]10)$ ]] && date -d@"$BASH_REMATCH[1]"; done <file.txt
      Wed Aug 24 20:09:55 BDT 2016
      Wed Aug 24 20:11:46 BDT 2016
      Wed Aug 24 20:13:58 BDT 2016

      $ while IFS= read -r i; do if [[ $i =~ ^#([0-9]10)$ ]]; then printf '#%sn' "$(date -d@"$BASH_REMATCH[1]")"; else printf '%sn' "$i"; fi; done <file.txt
      #Wed Aug 24 20:09:55 BDT 2016
      ll /data/holding/email
      #Wed Aug 24 20:11:46 BDT 2016
      cat /etc/rsyslog.conf
      #Wed Aug 24 20:13:58 BDT 2016
      ll /data/holding/web





      share|improve this answer














      Assuming consistent file format, with bash you can read the file line by line, test if it's in given format and then do the conversion:



      while IFS= read -r i; do [[ $i =~ ^#([0-9]10)$ ]] && 
      date -d@"$BASH_REMATCH[1]"; done <file.txt


      BASH_REMATCH is an array whose first element is the first captured group in Regex matching, =~, in this case the epoch.




      If you want to keep the file structure:



      while IFS= read -r i; do if [[ $i =~ ^#([0-9]10)$ ]]; then printf '#%sn' 
      "$(date -d@"$BASH_REMATCH[1]")"; else printf '%sn' "$i"; fi; done <file.txt


      this will output the modified contents to STDOUT, to save it in a file e.g. out.txt:



      while ...; do ...; done >out.txt


      Now if you wish, you can replace the original file:



      mv out.txt file.txt



      Example:



      $ cat file.txt
      #1472047795
      ll /data/holding/email
      #1472047906
      cat /etc/rsyslog.conf
      #1472048038
      ll /data/holding/web

      $ while IFS= read -r i; do [[ $i =~ ^#([0-9]10)$ ]] && date -d@"$BASH_REMATCH[1]"; done <file.txt
      Wed Aug 24 20:09:55 BDT 2016
      Wed Aug 24 20:11:46 BDT 2016
      Wed Aug 24 20:13:58 BDT 2016

      $ while IFS= read -r i; do if [[ $i =~ ^#([0-9]10)$ ]]; then printf '#%sn' "$(date -d@"$BASH_REMATCH[1]")"; else printf '%sn' "$i"; fi; done <file.txt
      #Wed Aug 24 20:09:55 BDT 2016
      ll /data/holding/email
      #Wed Aug 24 20:11:46 BDT 2016
      cat /etc/rsyslog.conf
      #Wed Aug 24 20:13:58 BDT 2016
      ll /data/holding/web






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Aug 30 '16 at 8:17

























      answered Aug 30 '16 at 7:53









      heemayl

      34k370100




      34k370100











      • Nice....that prints the converted date to screen, now how do I get that command to replace the entries in the file?
        – machinist
        Aug 30 '16 at 8:01










      • @machinist Check my edits..
        – heemayl
        Aug 30 '16 at 8:17






      • 1




        If you are using a recent version of bash, printf can do the conversion itself: printf '#%(%F %H)Tn' "$BASH_REMATCH[1]".
        – chepner
        Aug 30 '16 at 19:54

















      • Nice....that prints the converted date to screen, now how do I get that command to replace the entries in the file?
        – machinist
        Aug 30 '16 at 8:01










      • @machinist Check my edits..
        – heemayl
        Aug 30 '16 at 8:17






      • 1




        If you are using a recent version of bash, printf can do the conversion itself: printf '#%(%F %H)Tn' "$BASH_REMATCH[1]".
        – chepner
        Aug 30 '16 at 19:54
















      Nice....that prints the converted date to screen, now how do I get that command to replace the entries in the file?
      – machinist
      Aug 30 '16 at 8:01




      Nice....that prints the converted date to screen, now how do I get that command to replace the entries in the file?
      – machinist
      Aug 30 '16 at 8:01












      @machinist Check my edits..
      – heemayl
      Aug 30 '16 at 8:17




      @machinist Check my edits..
      – heemayl
      Aug 30 '16 at 8:17




      1




      1




      If you are using a recent version of bash, printf can do the conversion itself: printf '#%(%F %H)Tn' "$BASH_REMATCH[1]".
      – chepner
      Aug 30 '16 at 19:54





      If you are using a recent version of bash, printf can do the conversion itself: printf '#%(%F %H)Tn' "$BASH_REMATCH[1]".
      – chepner
      Aug 30 '16 at 19:54













      up vote
      14
      down vote













      While it's possible with GNU sed with things like:



      sed -E 's/^#([0-9]+).*$/date -d @1/e'


      That would be terribly inefficient (and is easy to introduce arbitrary command injection vulnerabilities1) as that would mean running one shell and one date command for each #xxxx line, virtually as bad as a shell while read loop. Here, it would be better to use things like perl or gawk, that is text processing utilities that have date conversion capabilities built-in:



      perl -MPOSIX -pe 's/^#(d+).*/ctime $1/se'


      Or:



      gawk '/^#/$0 = strftime("%c", substr($0, 2));1'



      1 If we had written ^#([0-9]).* instead of ^#([0-9]).*$ (as I did in an earlier version of this answer), then in multi-byte locales like UTF-8 ones (the norm nowadays), with an input like #1472047795<0x80>;reboot, where that <0x80> is the byte value 0x80 which does not form a valid character, that s command would have ended up running date -d@1472047795<0x80>; reboot for instance. While with the extra $, those lines would not be substituted. An alternative approach would be: s/^#([0-9])/date -d @1 #/e, that is leave the part after the #xxx date as a shell comment






      share|improve this answer






















      • What about using just a single instance of date -f to do all the conversions in a stream-wise manner?
        – Digital Trauma
        Aug 30 '16 at 20:58










      • The perl command seems to add a new line after ctime $1 and I can't find any way to remove it.
        – Alex Harvey
        Jun 23 '17 at 7:47







      • 1




        @Alex. Right. See edit. Adding the s flag makes so that .* also includes the newline on input. You can also use strftime "%c", localtime $1.
        – Stéphane Chazelas
        Jun 23 '17 at 14:02










      • @StéphaneChazelas thanks so much. It's a great answer.
        – Alex Harvey
        Jun 23 '17 at 14:04














      up vote
      14
      down vote













      While it's possible with GNU sed with things like:



      sed -E 's/^#([0-9]+).*$/date -d @1/e'


      That would be terribly inefficient (and is easy to introduce arbitrary command injection vulnerabilities1) as that would mean running one shell and one date command for each #xxxx line, virtually as bad as a shell while read loop. Here, it would be better to use things like perl or gawk, that is text processing utilities that have date conversion capabilities built-in:



      perl -MPOSIX -pe 's/^#(d+).*/ctime $1/se'


      Or:



      gawk '/^#/$0 = strftime("%c", substr($0, 2));1'



      1 If we had written ^#([0-9]).* instead of ^#([0-9]).*$ (as I did in an earlier version of this answer), then in multi-byte locales like UTF-8 ones (the norm nowadays), with an input like #1472047795<0x80>;reboot, where that <0x80> is the byte value 0x80 which does not form a valid character, that s command would have ended up running date -d@1472047795<0x80>; reboot for instance. While with the extra $, those lines would not be substituted. An alternative approach would be: s/^#([0-9])/date -d @1 #/e, that is leave the part after the #xxx date as a shell comment






      share|improve this answer






















      • What about using just a single instance of date -f to do all the conversions in a stream-wise manner?
        – Digital Trauma
        Aug 30 '16 at 20:58










      • The perl command seems to add a new line after ctime $1 and I can't find any way to remove it.
        – Alex Harvey
        Jun 23 '17 at 7:47







      • 1




        @Alex. Right. See edit. Adding the s flag makes so that .* also includes the newline on input. You can also use strftime "%c", localtime $1.
        – Stéphane Chazelas
        Jun 23 '17 at 14:02










      • @StéphaneChazelas thanks so much. It's a great answer.
        – Alex Harvey
        Jun 23 '17 at 14:04












      up vote
      14
      down vote










      up vote
      14
      down vote









      While it's possible with GNU sed with things like:



      sed -E 's/^#([0-9]+).*$/date -d @1/e'


      That would be terribly inefficient (and is easy to introduce arbitrary command injection vulnerabilities1) as that would mean running one shell and one date command for each #xxxx line, virtually as bad as a shell while read loop. Here, it would be better to use things like perl or gawk, that is text processing utilities that have date conversion capabilities built-in:



      perl -MPOSIX -pe 's/^#(d+).*/ctime $1/se'


      Or:



      gawk '/^#/$0 = strftime("%c", substr($0, 2));1'



      1 If we had written ^#([0-9]).* instead of ^#([0-9]).*$ (as I did in an earlier version of this answer), then in multi-byte locales like UTF-8 ones (the norm nowadays), with an input like #1472047795<0x80>;reboot, where that <0x80> is the byte value 0x80 which does not form a valid character, that s command would have ended up running date -d@1472047795<0x80>; reboot for instance. While with the extra $, those lines would not be substituted. An alternative approach would be: s/^#([0-9])/date -d @1 #/e, that is leave the part after the #xxx date as a shell comment






      share|improve this answer














      While it's possible with GNU sed with things like:



      sed -E 's/^#([0-9]+).*$/date -d @1/e'


      That would be terribly inefficient (and is easy to introduce arbitrary command injection vulnerabilities1) as that would mean running one shell and one date command for each #xxxx line, virtually as bad as a shell while read loop. Here, it would be better to use things like perl or gawk, that is text processing utilities that have date conversion capabilities built-in:



      perl -MPOSIX -pe 's/^#(d+).*/ctime $1/se'


      Or:



      gawk '/^#/$0 = strftime("%c", substr($0, 2));1'



      1 If we had written ^#([0-9]).* instead of ^#([0-9]).*$ (as I did in an earlier version of this answer), then in multi-byte locales like UTF-8 ones (the norm nowadays), with an input like #1472047795<0x80>;reboot, where that <0x80> is the byte value 0x80 which does not form a valid character, that s command would have ended up running date -d@1472047795<0x80>; reboot for instance. While with the extra $, those lines would not be substituted. An alternative approach would be: s/^#([0-9])/date -d @1 #/e, that is leave the part after the #xxx date as a shell comment







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Jun 23 '17 at 14:00

























      answered Aug 30 '16 at 8:33









      Stéphane Chazelas

      294k54552894




      294k54552894











      • What about using just a single instance of date -f to do all the conversions in a stream-wise manner?
        – Digital Trauma
        Aug 30 '16 at 20:58










      • The perl command seems to add a new line after ctime $1 and I can't find any way to remove it.
        – Alex Harvey
        Jun 23 '17 at 7:47







      • 1




        @Alex. Right. See edit. Adding the s flag makes so that .* also includes the newline on input. You can also use strftime "%c", localtime $1.
        – Stéphane Chazelas
        Jun 23 '17 at 14:02










      • @StéphaneChazelas thanks so much. It's a great answer.
        – Alex Harvey
        Jun 23 '17 at 14:04
















      • What about using just a single instance of date -f to do all the conversions in a stream-wise manner?
        – Digital Trauma
        Aug 30 '16 at 20:58










      • The perl command seems to add a new line after ctime $1 and I can't find any way to remove it.
        – Alex Harvey
        Jun 23 '17 at 7:47







      • 1




        @Alex. Right. See edit. Adding the s flag makes so that .* also includes the newline on input. You can also use strftime "%c", localtime $1.
        – Stéphane Chazelas
        Jun 23 '17 at 14:02










      • @StéphaneChazelas thanks so much. It's a great answer.
        – Alex Harvey
        Jun 23 '17 at 14:04















      What about using just a single instance of date -f to do all the conversions in a stream-wise manner?
      – Digital Trauma
      Aug 30 '16 at 20:58




      What about using just a single instance of date -f to do all the conversions in a stream-wise manner?
      – Digital Trauma
      Aug 30 '16 at 20:58












      The perl command seems to add a new line after ctime $1 and I can't find any way to remove it.
      – Alex Harvey
      Jun 23 '17 at 7:47





      The perl command seems to add a new line after ctime $1 and I can't find any way to remove it.
      – Alex Harvey
      Jun 23 '17 at 7:47





      1




      1




      @Alex. Right. See edit. Adding the s flag makes so that .* also includes the newline on input. You can also use strftime "%c", localtime $1.
      – Stéphane Chazelas
      Jun 23 '17 at 14:02




      @Alex. Right. See edit. Adding the s flag makes so that .* also includes the newline on input. You can also use strftime "%c", localtime $1.
      – Stéphane Chazelas
      Jun 23 '17 at 14:02












      @StéphaneChazelas thanks so much. It's a great answer.
      – Alex Harvey
      Jun 23 '17 at 14:04




      @StéphaneChazelas thanks so much. It's a great answer.
      – Alex Harvey
      Jun 23 '17 at 14:04










      up vote
      2
      down vote













      All the other answers spawn a new date process for every epoch date that needs to be converted. This could potentially add performance overhead if your input is large.



      However GNU date has a handy -f option that allows a single process instance of date to continuously read input dates without the need for a new fork. So we can use sed, paste and date in this manner such that each one only gets spawned once (2x for sed) regardless of how large the input is:



      $ paste -d 'n' <( sed '2~2d;y/#/@/' epoch.txt | date -f - ) <( sed '1~2d' epoch.txt )
      Wed Aug 24 07:09:55 PDT 2016
      ll /data/holding/email
      Wed Aug 24 07:11:46 PDT 2016
      cat /etc/rsyslog.conf
      Wed Aug 24 07:13:58 PDT 2016
      ll /data/holding/web
      $


      • The two sed commands respectively basically delete even and odd lines of the input; the first one also replaces # with @ to give the correct epoch timestamp format.

      • The first sed output is then piped to date -f which does the required date conversion, for every line of input that it receives.

      • These two streams are then interlaced into the single required output using paste. The <( ) constructs are bash process substitutions that effectively trick paste into thinking it is reading from given filenames when it is in fact reading the output piped from the command inside. -d 'n' tells paste to separate odd and even output lines with a newline. You could change (or remove) this if for example you want the timestamp on the same line as the other text.

      Note that there are several GNUisms and Bashisms in this command. This is not Posix-compliant and should not be expected to be portable outside of the GNU/Linux world. For example date -f does something else on OSXes BSD date variant.






      share|improve this answer




















      • date -d (from the question) is also non-portable... (On FreeBSD it will try to mess with DST settings, on Solaris it will give an error...) The question does not specify an OS though...
        – Gert van den Berg
        Jan 13 '17 at 8:14










      • @GertvandenBerg yes, this is addressed in the last paragraph of this answer.
        – Digital Trauma
        Jan 13 '17 at 15:56










      • I mean that the asker's sample also has portability issues... (They should probably have tagged an OS...)
        – Gert van den Berg
        Jan 14 '17 at 16:06














      up vote
      2
      down vote













      All the other answers spawn a new date process for every epoch date that needs to be converted. This could potentially add performance overhead if your input is large.



      However GNU date has a handy -f option that allows a single process instance of date to continuously read input dates without the need for a new fork. So we can use sed, paste and date in this manner such that each one only gets spawned once (2x for sed) regardless of how large the input is:



      $ paste -d 'n' <( sed '2~2d;y/#/@/' epoch.txt | date -f - ) <( sed '1~2d' epoch.txt )
      Wed Aug 24 07:09:55 PDT 2016
      ll /data/holding/email
      Wed Aug 24 07:11:46 PDT 2016
      cat /etc/rsyslog.conf
      Wed Aug 24 07:13:58 PDT 2016
      ll /data/holding/web
      $


      • The two sed commands respectively basically delete even and odd lines of the input; the first one also replaces # with @ to give the correct epoch timestamp format.

      • The first sed output is then piped to date -f which does the required date conversion, for every line of input that it receives.

      • These two streams are then interlaced into the single required output using paste. The <( ) constructs are bash process substitutions that effectively trick paste into thinking it is reading from given filenames when it is in fact reading the output piped from the command inside. -d 'n' tells paste to separate odd and even output lines with a newline. You could change (or remove) this if for example you want the timestamp on the same line as the other text.

      Note that there are several GNUisms and Bashisms in this command. This is not Posix-compliant and should not be expected to be portable outside of the GNU/Linux world. For example date -f does something else on OSXes BSD date variant.






      share|improve this answer




















      • date -d (from the question) is also non-portable... (On FreeBSD it will try to mess with DST settings, on Solaris it will give an error...) The question does not specify an OS though...
        – Gert van den Berg
        Jan 13 '17 at 8:14










      • @GertvandenBerg yes, this is addressed in the last paragraph of this answer.
        – Digital Trauma
        Jan 13 '17 at 15:56










      • I mean that the asker's sample also has portability issues... (They should probably have tagged an OS...)
        – Gert van den Berg
        Jan 14 '17 at 16:06












      up vote
      2
      down vote










      up vote
      2
      down vote









      All the other answers spawn a new date process for every epoch date that needs to be converted. This could potentially add performance overhead if your input is large.



      However GNU date has a handy -f option that allows a single process instance of date to continuously read input dates without the need for a new fork. So we can use sed, paste and date in this manner such that each one only gets spawned once (2x for sed) regardless of how large the input is:



      $ paste -d 'n' <( sed '2~2d;y/#/@/' epoch.txt | date -f - ) <( sed '1~2d' epoch.txt )
      Wed Aug 24 07:09:55 PDT 2016
      ll /data/holding/email
      Wed Aug 24 07:11:46 PDT 2016
      cat /etc/rsyslog.conf
      Wed Aug 24 07:13:58 PDT 2016
      ll /data/holding/web
      $


      • The two sed commands respectively basically delete even and odd lines of the input; the first one also replaces # with @ to give the correct epoch timestamp format.

      • The first sed output is then piped to date -f which does the required date conversion, for every line of input that it receives.

      • These two streams are then interlaced into the single required output using paste. The <( ) constructs are bash process substitutions that effectively trick paste into thinking it is reading from given filenames when it is in fact reading the output piped from the command inside. -d 'n' tells paste to separate odd and even output lines with a newline. You could change (or remove) this if for example you want the timestamp on the same line as the other text.

      Note that there are several GNUisms and Bashisms in this command. This is not Posix-compliant and should not be expected to be portable outside of the GNU/Linux world. For example date -f does something else on OSXes BSD date variant.






      share|improve this answer












      All the other answers spawn a new date process for every epoch date that needs to be converted. This could potentially add performance overhead if your input is large.



      However GNU date has a handy -f option that allows a single process instance of date to continuously read input dates without the need for a new fork. So we can use sed, paste and date in this manner such that each one only gets spawned once (2x for sed) regardless of how large the input is:



      $ paste -d 'n' <( sed '2~2d;y/#/@/' epoch.txt | date -f - ) <( sed '1~2d' epoch.txt )
      Wed Aug 24 07:09:55 PDT 2016
      ll /data/holding/email
      Wed Aug 24 07:11:46 PDT 2016
      cat /etc/rsyslog.conf
      Wed Aug 24 07:13:58 PDT 2016
      ll /data/holding/web
      $


      • The two sed commands respectively basically delete even and odd lines of the input; the first one also replaces # with @ to give the correct epoch timestamp format.

      • The first sed output is then piped to date -f which does the required date conversion, for every line of input that it receives.

      • These two streams are then interlaced into the single required output using paste. The <( ) constructs are bash process substitutions that effectively trick paste into thinking it is reading from given filenames when it is in fact reading the output piped from the command inside. -d 'n' tells paste to separate odd and even output lines with a newline. You could change (or remove) this if for example you want the timestamp on the same line as the other text.

      Note that there are several GNUisms and Bashisms in this command. This is not Posix-compliant and should not be expected to be portable outside of the GNU/Linux world. For example date -f does something else on OSXes BSD date variant.







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered Aug 30 '16 at 20:49









      Digital Trauma

      5,75211528




      5,75211528











      • date -d (from the question) is also non-portable... (On FreeBSD it will try to mess with DST settings, on Solaris it will give an error...) The question does not specify an OS though...
        – Gert van den Berg
        Jan 13 '17 at 8:14










      • @GertvandenBerg yes, this is addressed in the last paragraph of this answer.
        – Digital Trauma
        Jan 13 '17 at 15:56










      • I mean that the asker's sample also has portability issues... (They should probably have tagged an OS...)
        – Gert van den Berg
        Jan 14 '17 at 16:06
















      • date -d (from the question) is also non-portable... (On FreeBSD it will try to mess with DST settings, on Solaris it will give an error...) The question does not specify an OS though...
        – Gert van den Berg
        Jan 13 '17 at 8:14










      • @GertvandenBerg yes, this is addressed in the last paragraph of this answer.
        – Digital Trauma
        Jan 13 '17 at 15:56










      • I mean that the asker's sample also has portability issues... (They should probably have tagged an OS...)
        – Gert van den Berg
        Jan 14 '17 at 16:06















      date -d (from the question) is also non-portable... (On FreeBSD it will try to mess with DST settings, on Solaris it will give an error...) The question does not specify an OS though...
      – Gert van den Berg
      Jan 13 '17 at 8:14




      date -d (from the question) is also non-portable... (On FreeBSD it will try to mess with DST settings, on Solaris it will give an error...) The question does not specify an OS though...
      – Gert van den Berg
      Jan 13 '17 at 8:14












      @GertvandenBerg yes, this is addressed in the last paragraph of this answer.
      – Digital Trauma
      Jan 13 '17 at 15:56




      @GertvandenBerg yes, this is addressed in the last paragraph of this answer.
      – Digital Trauma
      Jan 13 '17 at 15:56












      I mean that the asker's sample also has portability issues... (They should probably have tagged an OS...)
      – Gert van den Berg
      Jan 14 '17 at 16:06




      I mean that the asker's sample also has portability issues... (They should probably have tagged an OS...)
      – Gert van den Berg
      Jan 14 '17 at 16:06










      up vote
      1
      down vote













      Assuming the date format you have in your post is what you want, the following regex should fit your needs.



      sed -E 's/#(1[0-9]9)(.*)/echo 1 $(date -d @1)/e' log.file


      Be mindful of the fact this will only replace one epoch per line.






      share|improve this answer






















      • I'm getting the following error with that command: sed: -e expression #1, char 48: invalid reference 3 on 's' command's RHS
        – machinist
        Aug 30 '16 at 7:53







      • 1




        My mistake, edited the post.
        – Hatclock
        Aug 30 '16 at 8:25














      up vote
      1
      down vote













      Assuming the date format you have in your post is what you want, the following regex should fit your needs.



      sed -E 's/#(1[0-9]9)(.*)/echo 1 $(date -d @1)/e' log.file


      Be mindful of the fact this will only replace one epoch per line.






      share|improve this answer






















      • I'm getting the following error with that command: sed: -e expression #1, char 48: invalid reference 3 on 's' command's RHS
        – machinist
        Aug 30 '16 at 7:53







      • 1




        My mistake, edited the post.
        – Hatclock
        Aug 30 '16 at 8:25












      up vote
      1
      down vote










      up vote
      1
      down vote









      Assuming the date format you have in your post is what you want, the following regex should fit your needs.



      sed -E 's/#(1[0-9]9)(.*)/echo 1 $(date -d @1)/e' log.file


      Be mindful of the fact this will only replace one epoch per line.






      share|improve this answer














      Assuming the date format you have in your post is what you want, the following regex should fit your needs.



      sed -E 's/#(1[0-9]9)(.*)/echo 1 $(date -d @1)/e' log.file


      Be mindful of the fact this will only replace one epoch per line.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Aug 30 '16 at 8:25

























      answered Aug 30 '16 at 7:48









      Hatclock

      656




      656











      • I'm getting the following error with that command: sed: -e expression #1, char 48: invalid reference 3 on 's' command's RHS
        – machinist
        Aug 30 '16 at 7:53







      • 1




        My mistake, edited the post.
        – Hatclock
        Aug 30 '16 at 8:25
















      • I'm getting the following error with that command: sed: -e expression #1, char 48: invalid reference 3 on 's' command's RHS
        – machinist
        Aug 30 '16 at 7:53







      • 1




        My mistake, edited the post.
        – Hatclock
        Aug 30 '16 at 8:25















      I'm getting the following error with that command: sed: -e expression #1, char 48: invalid reference 3 on 's' command's RHS
      – machinist
      Aug 30 '16 at 7:53





      I'm getting the following error with that command: sed: -e expression #1, char 48: invalid reference 3 on 's' command's RHS
      – machinist
      Aug 30 '16 at 7:53





      1




      1




      My mistake, edited the post.
      – Hatclock
      Aug 30 '16 at 8:25




      My mistake, edited the post.
      – Hatclock
      Aug 30 '16 at 8:25










      up vote
      0
      down vote













      using sed :



      sed -r 's/#([0-9]*)/echo $(date -d @1)/eg' test.txt


      output :



      ر أغس 24 16:09:55 EET 2016
      ll /data/holding/email
      ر أغس 24 16:11:46 EET 2016
      cat /etc/rsyslog.conf
      ر أغس 24 16:13:58 EET 2016
      ll /data/holding/web


      as my locale language is Arabic :)






      share|improve this answer


























        up vote
        0
        down vote













        using sed :



        sed -r 's/#([0-9]*)/echo $(date -d @1)/eg' test.txt


        output :



        ر أغس 24 16:09:55 EET 2016
        ll /data/holding/email
        ر أغس 24 16:11:46 EET 2016
        cat /etc/rsyslog.conf
        ر أغس 24 16:13:58 EET 2016
        ll /data/holding/web


        as my locale language is Arabic :)






        share|improve this answer
























          up vote
          0
          down vote










          up vote
          0
          down vote









          using sed :



          sed -r 's/#([0-9]*)/echo $(date -d @1)/eg' test.txt


          output :



          ر أغس 24 16:09:55 EET 2016
          ll /data/holding/email
          ر أغس 24 16:11:46 EET 2016
          cat /etc/rsyslog.conf
          ر أغس 24 16:13:58 EET 2016
          ll /data/holding/web


          as my locale language is Arabic :)






          share|improve this answer














          using sed :



          sed -r 's/#([0-9]*)/echo $(date -d @1)/eg' test.txt


          output :



          ر أغس 24 16:09:55 EET 2016
          ll /data/holding/email
          ر أغس 24 16:11:46 EET 2016
          cat /etc/rsyslog.conf
          ر أغس 24 16:13:58 EET 2016
          ll /data/holding/web


          as my locale language is Arabic :)







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Aug 30 '16 at 8:32

























          answered Aug 30 '16 at 8:07









          hassan

          17115




          17115




















              up vote
              0
              down vote













              My solution how to do that in a pipeline



              cat test.txt | sed 's/^/echo "/; s/([0-9]10)/`date -d @1`/; s/$/"/' | bash





              share|improve this answer
























                up vote
                0
                down vote













                My solution how to do that in a pipeline



                cat test.txt | sed 's/^/echo "/; s/([0-9]10)/`date -d @1`/; s/$/"/' | bash





                share|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  My solution how to do that in a pipeline



                  cat test.txt | sed 's/^/echo "/; s/([0-9]10)/`date -d @1`/; s/$/"/' | bash





                  share|improve this answer












                  My solution how to do that in a pipeline



                  cat test.txt | sed 's/^/echo "/; s/([0-9]10)/`date -d @1`/; s/$/"/' | bash






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Jan 13 '17 at 7:50









                  kayn

                  1312




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