how to capture specific strings from line

up vote
1
down vote

favorite

how to capture only the sdX from the following line ( withe bash/awk/sed/perl one liner )

echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data","

expected output

sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 12:02

FaxMax

423219

asked Dec 22 '17 at 9:48

yael

2,0091145

add a commentÂ |Â

up vote
1
down vote

favorite

how to capture only the sdX from the following line ( withe bash/awk/sed/perl one liner )

echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data","

expected output

sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 12:02

FaxMax

423219

asked Dec 22 '17 at 9:48

yael

2,0091145

add a commentÂ |Â

up vote
1
down vote

favorite

how to capture only the sdX from the following line ( withe bash/awk/sed/perl one liner )

echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data","

expected output

sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 12:02

FaxMax

423219

asked Dec 22 '17 at 9:48

yael

2,0091145

how to capture only the sdX from the following line ( withe bash/awk/sed/perl one liner )

echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data","

expected output

sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 12:02

FaxMax

423219

asked Dec 22 '17 at 9:48

yael

2,0091145

edited Dec 22 '17 at 12:02

FaxMax

423219

edited Dec 22 '17 at 12:02

FaxMax

423219

edited Dec 22 '17 at 12:02

FaxMax

423219

asked Dec 22 '17 at 9:48

yael

2,0091145

asked Dec 22 '17 at 9:48

yael

2,0091145

asked Dec 22 '17 at 9:48

yael

2,0091145

add a commentÂ |Â

5 Answers
5

active

oldest

votes

up vote
2
down vote

accepted

You can use GREP arguments:

 -P, --perl-regexp
 Interpret the pattern as a Perl-compatible regular expression
 (PCRE). This is experimental and grep -P may warn of
 unimplemented features.
 -o, --only-matching
 Print only the matched (non-empty) parts of a matching line,
 with each such part on a separate output line.

So your command would be:

echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data"," | grep -oP "w*sdw*"
sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 11:04

answered Dec 22 '17 at 10:41

Kevin Lemaire

1,037421

1

Sorry, you talk about the option -P and the you use -h to hide the filename? grep is (in this case) reading from stdin
â€“Â FaxMax
Dec 22 '17 at 10:48

Corrected, wrong copy/paste. Thanks.
â€“Â Kevin Lemaire
Dec 22 '17 at 10:52

add a commentÂ |Â

up vote
1
down vote

Use

echo ... | grep -Eo "sd[a-z]"

where -E interprets the pattern as an (extended) regular expression and -o prints only the matching parts in each line.

answered Dec 22 '17 at 10:48

elm

1134

add a commentÂ |Â

up vote
0
down vote

You did not specify what x should be in sdX so one could as well assume that it might be any character:

$ echo echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs /data"," | grep -o 'sd.'
sdb
sdc
sdd
sde
sdf

A literal . means any character in regular expressions.

And in perl because you wanted that explicitly:

$ echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data"," | perl -nE 'say $& while /sd./g'
sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 12:15

answered Dec 22 '17 at 11:03

Arkadiusz Drabczyk

7,21521532

add a commentÂ |Â

up vote
0
down vote

Above result is achieved by awk one liner

 echo " ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data","" | sed "s/,/n/g" | awk -F "/" 'print $3' | sed '/^$/d'

output

sdb
sdc
sdd
sde
sdf

answered Dec 23 '17 at 9:54

Praveen Kumar BS

1,010128

add a commentÂ |Â

up vote
-1
down vote

echo '"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",' 
"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",
echo '"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",' | grep -Po 'sdw'
sdb
sdc
sdd
sde
sdf

answered Dec 22 '17 at 10:37

FaxMax

423219

add a commentÂ |Â

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
2
down vote

accepted

You can use GREP arguments:

 -P, --perl-regexp
 Interpret the pattern as a Perl-compatible regular expression
 (PCRE). This is experimental and grep -P may warn of
 unimplemented features.
 -o, --only-matching
 Print only the matched (non-empty) parts of a matching line,
 with each such part on a separate output line.

So your command would be:

echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data"," | grep -oP "w*sdw*"
sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 11:04

answered Dec 22 '17 at 10:41

Kevin Lemaire

1,037421

1

Sorry, you talk about the option -P and the you use -h to hide the filename? grep is (in this case) reading from stdin
â€“Â FaxMax
Dec 22 '17 at 10:48

Corrected, wrong copy/paste. Thanks.
â€“Â Kevin Lemaire
Dec 22 '17 at 10:52

add a commentÂ |Â

up vote
2
down vote

accepted

You can use GREP arguments:

 -P, --perl-regexp
 Interpret the pattern as a Perl-compatible regular expression
 (PCRE). This is experimental and grep -P may warn of
 unimplemented features.
 -o, --only-matching
 Print only the matched (non-empty) parts of a matching line,
 with each such part on a separate output line.

So your command would be:

echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data"," | grep -oP "w*sdw*"
sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 11:04

answered Dec 22 '17 at 10:41

Kevin Lemaire

1,037421

1

Sorry, you talk about the option -P and the you use -h to hide the filename? grep is (in this case) reading from stdin
â€“Â FaxMax
Dec 22 '17 at 10:48

Corrected, wrong copy/paste. Thanks.
â€“Â Kevin Lemaire
Dec 22 '17 at 10:52

add a commentÂ |Â

up vote
2
down vote

accepted

You can use GREP arguments:

 -P, --perl-regexp
 Interpret the pattern as a Perl-compatible regular expression
 (PCRE). This is experimental and grep -P may warn of
 unimplemented features.
 -o, --only-matching
 Print only the matched (non-empty) parts of a matching line,
 with each such part on a separate output line.

So your command would be:

echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data"," | grep -oP "w*sdw*"
sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 11:04

answered Dec 22 '17 at 10:41

Kevin Lemaire

1,037421

You can use GREP arguments:

 -P, --perl-regexp
 Interpret the pattern as a Perl-compatible regular expression
 (PCRE). This is experimental and grep -P may warn of
 unimplemented features.
 -o, --only-matching
 Print only the matched (non-empty) parts of a matching line,
 with each such part on a separate output line.

So your command would be:

echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data"," | grep -oP "w*sdw*"
sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 11:04

answered Dec 22 '17 at 10:41

Kevin Lemaire

1,037421

edited Dec 22 '17 at 11:04

answered Dec 22 '17 at 10:41

Kevin Lemaire

1,037421

answered Dec 22 '17 at 10:41

Kevin Lemaire

1,037421

answered Dec 22 '17 at 10:41

Kevin Lemaire

1,037421

1

Sorry, you talk about the option -P and the you use -h to hide the filename? grep is (in this case) reading from stdin
â€“Â FaxMax
Dec 22 '17 at 10:48

Corrected, wrong copy/paste. Thanks.
â€“Â Kevin Lemaire
Dec 22 '17 at 10:52

add a commentÂ |Â

1

Sorry, you talk about the option -P and the you use -h to hide the filename? grep is (in this case) reading from stdin
â€“Â FaxMax
Dec 22 '17 at 10:48

Corrected, wrong copy/paste. Thanks.
â€“Â Kevin Lemaire
Dec 22 '17 at 10:52

Sorry, you talk about the option -P and the you use -h to hide the filename? grep is (in this case) reading from stdin
â€“Â FaxMax
Dec 22 '17 at 10:48

Corrected, wrong copy/paste. Thanks.
â€“Â Kevin Lemaire
Dec 22 '17 at 10:52

add a commentÂ |Â

up vote
1
down vote

Use

echo ... | grep -Eo "sd[a-z]"

where -E interprets the pattern as an (extended) regular expression and -o prints only the matching parts in each line.

answered Dec 22 '17 at 10:48

elm

1134

add a commentÂ |Â

up vote
1
down vote

Use

echo ... | grep -Eo "sd[a-z]"

where -E interprets the pattern as an (extended) regular expression and -o prints only the matching parts in each line.

answered Dec 22 '17 at 10:48

elm

1134

add a commentÂ |Â

up vote
1
down vote

Use

echo ... | grep -Eo "sd[a-z]"

where -E interprets the pattern as an (extended) regular expression and -o prints only the matching parts in each line.

answered Dec 22 '17 at 10:48

elm

1134

Use

echo ... | grep -Eo "sd[a-z]"

where -E interprets the pattern as an (extended) regular expression and -o prints only the matching parts in each line.

answered Dec 22 '17 at 10:48

elm

1134

answered Dec 22 '17 at 10:48

elm

1134

answered Dec 22 '17 at 10:48

elm

1134

answered Dec 22 '17 at 10:48

elm

1134

add a commentÂ |Â

up vote
0
down vote

You did not specify what x should be in sdX so one could as well assume that it might be any character:

$ echo echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs /data"," | grep -o 'sd.'
sdb
sdc
sdd
sde
sdf

A literal . means any character in regular expressions.

And in perl because you wanted that explicitly:

$ echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data"," | perl -nE 'say $& while /sd./g'
sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 12:15

answered Dec 22 '17 at 11:03

Arkadiusz Drabczyk

7,21521532

add a commentÂ |Â

up vote
0
down vote

You did not specify what x should be in sdX so one could as well assume that it might be any character:

$ echo echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs /data"," | grep -o 'sd.'
sdb
sdc
sdd
sde
sdf

A literal . means any character in regular expressions.

And in perl because you wanted that explicitly:

$ echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data"," | perl -nE 'say $& while /sd./g'
sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 12:15

answered Dec 22 '17 at 11:03

Arkadiusz Drabczyk

7,21521532

add a commentÂ |Â

up vote
0
down vote

You did not specify what x should be in sdX so one could as well assume that it might be any character:

$ echo echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs /data"," | grep -o 'sd.'
sdb
sdc
sdd
sde
sdf

A literal . means any character in regular expressions.

And in perl because you wanted that explicitly:

$ echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data"," | perl -nE 'say $& while /sd./g'
sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 12:15

answered Dec 22 '17 at 11:03

Arkadiusz Drabczyk

7,21521532

You did not specify what x should be in sdX so one could as well assume that it might be any character:

$ echo echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs /data"," | grep -o 'sd.'
sdb
sdc
sdd
sde
sdf

A literal . means any character in regular expressions.

And in perl because you wanted that explicitly:

$ echo ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data"," | perl -nE 'say $& while /sd./g'
sdb
sdc
sdd
sde
sdf

edited Dec 22 '17 at 12:15

answered Dec 22 '17 at 11:03

Arkadiusz Drabczyk

7,21521532

edited Dec 22 '17 at 12:15

answered Dec 22 '17 at 11:03

Arkadiusz Drabczyk

7,21521532

answered Dec 22 '17 at 11:03

Arkadiusz Drabczyk

7,21521532

answered Dec 22 '17 at 11:03

Arkadiusz Drabczyk

7,21521532

add a commentÂ |Â

up vote
0
down vote

Above result is achieved by awk one liner

 echo " ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data","" | sed "s/,/n/g" | awk -F "/" 'print $3' | sed '/^$/d'

output

sdb
sdc
sdd
sde
sdf

answered Dec 23 '17 at 9:54

Praveen Kumar BS

1,010128

add a commentÂ |Â

up vote
0
down vote

Above result is achieved by awk one liner

 echo " ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data","" | sed "s/,/n/g" | awk -F "/" 'print $3' | sed '/^$/d'

output

sdb
sdc
sdd
sde
sdf

answered Dec 23 '17 at 9:54

Praveen Kumar BS

1,010128

add a commentÂ |Â

up vote
0
down vote

Above result is achieved by awk one liner

 echo " ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data","" | sed "s/,/n/g" | awk -F "/" 'print $3' | sed '/^$/d'

output

sdb
sdc
sdd
sde
sdf

answered Dec 23 '17 at 9:54

Praveen Kumar BS

1,010128

Above result is achieved by awk one liner

 echo " ""dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data","" | sed "s/,/n/g" | awk -F "/" 'print $3' | sed '/^$/d'

output

sdb
sdc
sdd
sde
sdf

answered Dec 23 '17 at 9:54

Praveen Kumar BS

1,010128

answered Dec 23 '17 at 9:54

Praveen Kumar BS

1,010128

answered Dec 23 '17 at 9:54

Praveen Kumar BS

1,010128

answered Dec 23 '17 at 9:54

Praveen Kumar BS

1,010128

add a commentÂ |Â

up vote
-1
down vote

echo '"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",' 
"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",
echo '"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",' | grep -Po 'sdw'
sdb
sdc
sdd
sde
sdf

answered Dec 22 '17 at 10:37

FaxMax

423219

add a commentÂ |Â

up vote
-1
down vote

echo '"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",' 
"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",
echo '"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",' | grep -Po 'sdw'
sdb
sdc
sdd
sde
sdf

answered Dec 22 '17 at 10:37

FaxMax

423219

add a commentÂ |Â

up vote
-1
down vote

echo '"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",' 
"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",
echo '"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",' | grep -Po 'sdw'
sdb
sdc
sdd
sde
sdf

answered Dec 22 '17 at 10:37

FaxMax

423219

echo '"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",' 
"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",
echo '"dfs.datanode.data.dir" : "/rid/sdb/oo/hdfs/data,/rid/sdc/oo/hdfs/data,/rid/sdd/oo/hdfs/data,/rid/sde/oo/hdfs/data,/rid/sdf/oo/hdfs/data",' | grep -Po 'sdw'
sdb
sdc
sdd
sde
sdf

answered Dec 22 '17 at 10:37

FaxMax

423219

answered Dec 22 '17 at 10:37

FaxMax

423219

answered Dec 22 '17 at 10:37

FaxMax

423219

answered Dec 22 '17 at 10:37

FaxMax

423219

add a commentÂ |Â

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