Shell - Find the latest file which matches a given pattern
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I have the following files.
root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver*
-rw-r--r-- 1 root root 0 Jul 5 18:54 /client/folder299/version_1
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder300/ifolder1/version_b
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder301/ifolder2/version_c
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder302/ifolder3/version_d
-rw-r--r-- 1 root root 0 Jul 19 13:35 /client/folder300/version_2
-rw-r--r-- 1 root root 0 Jul 19 13:35 /client/folder301/version_3
-rw-r--r-- 1 root root 0 Jul 19 13:36 /client/folder302/version_4
I am trying to get the latest version file for a pattern matching an ID. Example is shown below.
root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver* | grep 299
-rw-r--r-- 1 root root 0 Jul 5 18:54 /client/folder299/version_1
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a
The latest version is version_a in the above example.
root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver* | grep 299 | tail -1
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a
I am told that this approach is not good to find a file (Why *not* parse `ls`? )and am looking for an alternative way like https://stackoverflow.com/a/26766782/9316558. Please let me know if something is not clear.
Update:
From below answer by Jasen, I could get the latest file in the path /client
find /client -path "*299*" -printf "%T@ %Pn" | sort -n | tail -1
But, the above command gives the latest file. I am looking for finding the latest version file.
bash shell files solaris date
asked Jul 19 at 8:45
Raj
36
 |Â
show 3 more comments
up vote
0
down vote
favorite
1
I have the following files.
root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver*
-rw-r--r-- 1 root root 0 Jul 5 18:54 /client/folder299/version_1
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder300/ifolder1/version_b
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder301/ifolder2/version_c
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder302/ifolder3/version_d
-rw-r--r-- 1 root root 0 Jul 19 13:35 /client/folder300/version_2
-rw-r--r-- 1 root root 0 Jul 19 13:35 /client/folder301/version_3
-rw-r--r-- 1 root root 0 Jul 19 13:36 /client/folder302/version_4
I am trying to get the latest version file for a pattern matching an ID. Example is shown below.
root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver* | grep 299
-rw-r--r-- 1 root root 0 Jul 5 18:54 /client/folder299/version_1
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a
The latest version is version_a in the above example.
root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver* | grep 299 | tail -1
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a
I am told that this approach is not good to find a file (Why *not* parse `ls`? )and am looking for an alternative way like https://stackoverflow.com/a/26766782/9316558. Please let me know if something is not clear.
Update:
From below answer by Jasen, I could get the latest file in the path /client
find /client -path "*299*" -printf "%T@ %Pn" | sort -n | tail -1
But, the above command gives the latest file. I am looking for finding the latest version file.
bash shell files solaris date
asked Jul 19 at 8:45
Raj
36
Will the latest version file always be the one that sorts last, or should file timestamps be used, or some other scheme?
– Kusalananda
Jul 19 at 8:55
@Kusalananda Yes, the latest version file will always be the one that sorts last. I mean the one that is created recently.
– Raj
Jul 19 at 9:29
Sorry I should have been more specific. Sorts last, by name? ... taking into account thatversion_3sorts afterversion_10.
– Kusalananda
Jul 19 at 9:30
@Raj: Does it have to be bash? If you would write your script in Zsh, you could do for instance afiles=( /client/*299*/ver*(Nom) )to get an array of files sorted in ascending order by modification time (o means sorting, m means modification time, N causes an empty array to be generated if no matching files exist). If you need descending order, use(^Nom)instead.
– user1934428
Jul 19 at 9:37
1
Please re-tag your Question to Solaris, as it's apparent from comments that you're only connecting from Linux, then runningfindon Solaris.
– Jeff Schaller
Jul 19 at 19:24
 |Â
show 3 more comments
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
I have the following files.
root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver*
-rw-r--r-- 1 root root 0 Jul 5 18:54 /client/folder299/version_1
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder300/ifolder1/version_b
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder301/ifolder2/version_c
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder302/ifolder3/version_d
-rw-r--r-- 1 root root 0 Jul 19 13:35 /client/folder300/version_2
-rw-r--r-- 1 root root 0 Jul 19 13:35 /client/folder301/version_3
-rw-r--r-- 1 root root 0 Jul 19 13:36 /client/folder302/version_4
I am trying to get the latest version file for a pattern matching an ID. Example is shown below.
root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver* | grep 299
-rw-r--r-- 1 root root 0 Jul 5 18:54 /client/folder299/version_1
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a
The latest version is version_a in the above example.
root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver* | grep 299 | tail -1
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a
I am told that this approach is not good to find a file (Why *not* parse `ls`? )and am looking for an alternative way like https://stackoverflow.com/a/26766782/9316558. Please let me know if something is not clear.
Update:
From below answer by Jasen, I could get the latest file in the path /client
find /client -path "*299*" -printf "%T@ %Pn" | sort -n | tail -1
But, the above command gives the latest file. I am looking for finding the latest version file.
bash shell files solaris date
asked Jul 19 at 8:45
Raj
36
I have the following files.
root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver*
-rw-r--r-- 1 root root 0 Jul 5 18:54 /client/folder299/version_1
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder300/ifolder1/version_b
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder301/ifolder2/version_c
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder302/ifolder3/version_d
-rw-r--r-- 1 root root 0 Jul 19 13:35 /client/folder300/version_2
-rw-r--r-- 1 root root 0 Jul 19 13:35 /client/folder301/version_3
-rw-r--r-- 1 root root 0 Jul 19 13:36 /client/folder302/version_4
I am trying to get the latest version file for a pattern matching an ID. Example is shown below.
root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver* | grep 299
-rw-r--r-- 1 root root 0 Jul 5 18:54 /client/folder299/version_1
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a
The latest version is version_a in the above example.
root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver* | grep 299 | tail -1
-rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a
I am told that this approach is not good to find a file (Why *not* parse `ls`? )and am looking for an alternative way like https://stackoverflow.com/a/26766782/9316558. Please let me know if something is not clear.
Update:
From below answer by Jasen, I could get the latest file in the path /client
find /client -path "*299*" -printf "%T@ %Pn" | sort -n | tail -1
But, the above command gives the latest file. I am looking for finding the latest version file.
bash shell files solaris date
asked Jul 19 at 8:45
Raj
36
edited Jul 24 at 10:33
asked Jul 19 at 8:45
Raj
36
asked Jul 19 at 8:45
Raj
36
asked Jul 19 at 8:45
Raj
36
36
Will the latest version file always be the one that sorts last, or should file timestamps be used, or some other scheme?
– Kusalananda
Jul 19 at 8:55
@Kusalananda Yes, the latest version file will always be the one that sorts last. I mean the one that is created recently.
– Raj
Jul 19 at 9:29
Sorry I should have been more specific. Sorts last, by name? ... taking into account thatversion_3sorts afterversion_10.
– Kusalananda
Jul 19 at 9:30
@Raj: Does it have to be bash? If you would write your script in Zsh, you could do for instance afiles=( /client/*299*/ver*(Nom) )to get an array of files sorted in ascending order by modification time (o means sorting, m means modification time, N causes an empty array to be generated if no matching files exist). If you need descending order, use(^Nom)instead.
– user1934428
Jul 19 at 9:37
1
Please re-tag your Question to Solaris, as it's apparent from comments that you're only connecting from Linux, then runningfindon Solaris.
– Jeff Schaller
Jul 19 at 19:24
 |Â
show 3 more comments
Will the latest version file always be the one that sorts last, or should file timestamps be used, or some other scheme?
– Kusalananda
Jul 19 at 8:55
@Kusalananda Yes, the latest version file will always be the one that sorts last. I mean the one that is created recently.
– Raj
Jul 19 at 9:29
Sorry I should have been more specific. Sorts last, by name? ... taking into account thatversion_3sorts afterversion_10.
– Kusalananda
Jul 19 at 9:30
@Raj: Does it have to be bash? If you would write your script in Zsh, you could do for instance afiles=( /client/*299*/ver*(Nom) )to get an array of files sorted in ascending order by modification time (o means sorting, m means modification time, N causes an empty array to be generated if no matching files exist). If you need descending order, use(^Nom)instead.
– user1934428
Jul 19 at 9:37
1
Please re-tag your Question to Solaris, as it's apparent from comments that you're only connecting from Linux, then runningfindon Solaris.
– Jeff Schaller
Jul 19 at 19:24
Will the latest version file always be the one that sorts last, or should file timestamps be used, or some other scheme?
– Kusalananda
Jul 19 at 8:55
Will the latest version file always be the one that sorts last, or should file timestamps be used, or some other scheme?
– Kusalananda
Jul 19 at 8:55
@Kusalananda Yes, the latest version file will always be the one that sorts last. I mean the one that is created recently.
– Raj
Jul 19 at 9:29
@Kusalananda Yes, the latest version file will always be the one that sorts last. I mean the one that is created recently.
– Raj
Jul 19 at 9:29
Sorry I should have been more specific. Sorts last, by name? ... taking into account that
version_3 sorts after version_10.– Kusalananda
Jul 19 at 9:30
Sorry I should have been more specific. Sorts last, by name? ... taking into account that
version_3 sorts after version_10.– Kusalananda
Jul 19 at 9:30
@Raj: Does it have to be bash? If you would write your script in Zsh, you could do for instance a
files=( /client/*299*/ver*(Nom) ) to get an array of files sorted in ascending order by modification time (o means sorting, m means modification time, N causes an empty array to be generated if no matching files exist). If you need descending order, use (^Nom) instead.– user1934428
Jul 19 at 9:37
@Raj: Does it have to be bash? If you would write your script in Zsh, you could do for instance a
files=( /client/*299*/ver*(Nom) ) to get an array of files sorted in ascending order by modification time (o means sorting, m means modification time, N causes an empty array to be generated if no matching files exist). If you need descending order, use (^Nom) instead.– user1934428
Jul 19 at 9:37
1
1
Please re-tag your Question to Solaris, as it's apparent from comments that you're only connecting from Linux, then running
find on Solaris.– Jeff Schaller
Jul 19 at 19:24
Please re-tag your Question to Solaris, as it's apparent from comments that you're only connecting from Linux, then running
find on Solaris.– Jeff Schaller
Jul 19 at 19:24
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
you can combine find and sort
find -path "some pattern" -printf "%T@ %Pn" | sort -n | tail -1
answered Jul 19 at 10:24
Jasen
1,925713
Sorry. Should I be running the command like this?find -name "299" -printf "%T@ %Pn" | sort -n | tail -1
– Raj
Jul 19 at 10:49
possibly-path "*299*"it understands the normal shell wildcards
– Jasen
Jul 19 at 10:50
it turns out-namewas wrong and you need-pathth match on directory names. answer edited.
– Jasen
Jul 19 at 10:54
yeah, you can add a start location before-pathif you want search other than the current directory, see the man page for find - there's many other options.
– Jasen
Jul 19 at 11:16
Thanks for the update. The command seems to work if executed from inside the folder where I need this value from.root@RV-VMBOX:/client# find -path "*299*" -printf "%T@ %Pn" | sort -n | tail -11530797283.5560000000 ifolder299/ifolder/version_aIn my case, I need to execute this command remotely as a user of the system. So, I understand that this will search in the home directory of the above user, where this folder doesn't exist. Is there a way in which we can extend the -path's argument to be more precise with the path?
– Raj
Jul 19 at 11:18
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
you can combine find and sort
find -path "some pattern" -printf "%T@ %Pn" | sort -n | tail -1
answered Jul 19 at 10:24
Jasen
1,925713
Sorry. Should I be running the command like this?find -name "299" -printf "%T@ %Pn" | sort -n | tail -1
– Raj
Jul 19 at 10:49
possibly-path "*299*"it understands the normal shell wildcards
– Jasen
Jul 19 at 10:50
it turns out-namewas wrong and you need-pathth match on directory names. answer edited.
– Jasen
Jul 19 at 10:54
yeah, you can add a start location before-pathif you want search other than the current directory, see the man page for find - there's many other options.
– Jasen
Jul 19 at 11:16
Thanks for the update. The command seems to work if executed from inside the folder where I need this value from.root@RV-VMBOX:/client# find -path "*299*" -printf "%T@ %Pn" | sort -n | tail -11530797283.5560000000 ifolder299/ifolder/version_aIn my case, I need to execute this command remotely as a user of the system. So, I understand that this will search in the home directory of the above user, where this folder doesn't exist. Is there a way in which we can extend the -path's argument to be more precise with the path?
– Raj
Jul 19 at 11:18
 |Â
show 4 more comments
up vote
1
down vote
you can combine find and sort
find -path "some pattern" -printf "%T@ %Pn" | sort -n | tail -1
answered Jul 19 at 10:24
Jasen
1,925713
Sorry. Should I be running the command like this?find -name "299" -printf "%T@ %Pn" | sort -n | tail -1
– Raj
Jul 19 at 10:49
possibly-path "*299*"it understands the normal shell wildcards
– Jasen
Jul 19 at 10:50
it turns out-namewas wrong and you need-pathth match on directory names. answer edited.
– Jasen
Jul 19 at 10:54
yeah, you can add a start location before-pathif you want search other than the current directory, see the man page for find - there's many other options.
– Jasen
Jul 19 at 11:16
Thanks for the update. The command seems to work if executed from inside the folder where I need this value from.root@RV-VMBOX:/client# find -path "*299*" -printf "%T@ %Pn" | sort -n | tail -11530797283.5560000000 ifolder299/ifolder/version_aIn my case, I need to execute this command remotely as a user of the system. So, I understand that this will search in the home directory of the above user, where this folder doesn't exist. Is there a way in which we can extend the -path's argument to be more precise with the path?
– Raj
Jul 19 at 11:18
 |Â
show 4 more comments
up vote
1
down vote
up vote
1
down vote
you can combine find and sort
find -path "some pattern" -printf "%T@ %Pn" | sort -n | tail -1
answered Jul 19 at 10:24
Jasen
1,925713
you can combine find and sort
find -path "some pattern" -printf "%T@ %Pn" | sort -n | tail -1
answered Jul 19 at 10:24
Jasen
1,925713
edited Jul 19 at 10:54
answered Jul 19 at 10:24
Jasen
1,925713
answered Jul 19 at 10:24
Jasen
1,925713
answered Jul 19 at 10:24
Jasen
1,925713
1,925713
Sorry. Should I be running the command like this?find -name "299" -printf "%T@ %Pn" | sort -n | tail -1
– Raj
Jul 19 at 10:49
possibly-path "*299*"it understands the normal shell wildcards
– Jasen
Jul 19 at 10:50
it turns out-namewas wrong and you need-pathth match on directory names. answer edited.
– Jasen
Jul 19 at 10:54
yeah, you can add a start location before-pathif you want search other than the current directory, see the man page for find - there's many other options.
– Jasen
Jul 19 at 11:16
Thanks for the update. The command seems to work if executed from inside the folder where I need this value from.root@RV-VMBOX:/client# find -path "*299*" -printf "%T@ %Pn" | sort -n | tail -11530797283.5560000000 ifolder299/ifolder/version_aIn my case, I need to execute this command remotely as a user of the system. So, I understand that this will search in the home directory of the above user, where this folder doesn't exist. Is there a way in which we can extend the -path's argument to be more precise with the path?
– Raj
Jul 19 at 11:18
 |Â
show 4 more comments
Sorry. Should I be running the command like this?find -name "299" -printf "%T@ %Pn" | sort -n | tail -1
– Raj
Jul 19 at 10:49
possibly-path "*299*"it understands the normal shell wildcards
– Jasen
Jul 19 at 10:50
it turns out-namewas wrong and you need-pathth match on directory names. answer edited.
– Jasen
Jul 19 at 10:54
yeah, you can add a start location before-pathif you want search other than the current directory, see the man page for find - there's many other options.
– Jasen
Jul 19 at 11:16
Thanks for the update. The command seems to work if executed from inside the folder where I need this value from.root@RV-VMBOX:/client# find -path "*299*" -printf "%T@ %Pn" | sort -n | tail -11530797283.5560000000 ifolder299/ifolder/version_aIn my case, I need to execute this command remotely as a user of the system. So, I understand that this will search in the home directory of the above user, where this folder doesn't exist. Is there a way in which we can extend the -path's argument to be more precise with the path?
– Raj
Jul 19 at 11:18
Sorry. Should I be running the command like this?
find -name "299" -printf "%T@ %Pn" | sort -n | tail -1– Raj
Jul 19 at 10:49
Sorry. Should I be running the command like this?
find -name "299" -printf "%T@ %Pn" | sort -n | tail -1– Raj
Jul 19 at 10:49
possibly
-path "*299*" it understands the normal shell wildcards– Jasen
Jul 19 at 10:50
possibly
-path "*299*" it understands the normal shell wildcards– Jasen
Jul 19 at 10:50
it turns out
-name was wrong and you need -path th match on directory names. answer edited.– Jasen
Jul 19 at 10:54
it turns out
-name was wrong and you need -path th match on directory names. answer edited.– Jasen
Jul 19 at 10:54
yeah, you can add a start location before
-path if you want search other than the current directory, see the man page for find - there's many other options.– Jasen
Jul 19 at 11:16
yeah, you can add a start location before
-path if you want search other than the current directory, see the man page for find - there's many other options.– Jasen
Jul 19 at 11:16
Thanks for the update. The command seems to work if executed from inside the folder where I need this value from.
root@RV-VMBOX:/client# find -path "*299*" -printf "%T@ %Pn" | sort -n | tail -1 1530797283.5560000000 ifolder299/ifolder/version_a In my case, I need to execute this command remotely as a user of the system. So, I understand that this will search in the home directory of the above user, where this folder doesn't exist. Is there a way in which we can extend the -path's argument to be more precise with the path?– Raj
Jul 19 at 11:18
Thanks for the update. The command seems to work if executed from inside the folder where I need this value from.
root@RV-VMBOX:/client# find -path "*299*" -printf "%T@ %Pn" | sort -n | tail -1 1530797283.5560000000 ifolder299/ifolder/version_a In my case, I need to execute this command remotely as a user of the system. So, I understand that this will search in the home directory of the above user, where this folder doesn't exist. Is there a way in which we can extend the -path's argument to be more precise with the path?– Raj
Jul 19 at 11:18
 |Â
show 4 more comments
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Will the latest version file always be the one that sorts last, or should file timestamps be used, or some other scheme?
– Kusalananda
Jul 19 at 8:55
@Kusalananda Yes, the latest version file will always be the one that sorts last. I mean the one that is created recently.
– Raj
Jul 19 at 9:29
Sorry I should have been more specific. Sorts last, by name? ... taking into account that
version_3sorts afterversion_10.– Kusalananda
Jul 19 at 9:30
@Raj: Does it have to be bash? If you would write your script in Zsh, you could do for instance a
files=( /client/*299*/ver*(Nom) )to get an array of files sorted in ascending order by modification time (o means sorting, m means modification time, N causes an empty array to be generated if no matching files exist). If you need descending order, use(^Nom)instead.– user1934428
Jul 19 at 9:37
1
Please re-tag your Question to Solaris, as it's apparent from comments that you're only connecting from Linux, then running
findon Solaris.– Jeff Schaller
Jul 19 at 19:24