Can not extract the capture group with neither sed nor grep
28
I want to extract the value pair from a key-value pair syntax but I can not.
Example I tried:
echo employee_id=1234 | sed 's/employee_id=([0-9]+)/1/g'
But this gives employee_id=1234 and not 1234 which is actually the capture group.
What am I doing wrong here? I also tried:
echo employee_id=1234| egrep -o employee_id=([0-9]+)
but no sucess.
regex linux sed grep
edited Sep 19 '13 at 18:41
Adrian Frühwirth
26.1k75163
asked Sep 19 '13 at 10:52
JimJim
7,5012495179
add a comment |
28
I want to extract the value pair from a key-value pair syntax but I can not.
Example I tried:
echo employee_id=1234 | sed 's/employee_id=([0-9]+)/1/g'
But this gives employee_id=1234 and not 1234 which is actually the capture group.
What am I doing wrong here? I also tried:
echo employee_id=1234| egrep -o employee_id=([0-9]+)
but no sucess.
regex linux sed grep
edited Sep 19 '13 at 18:41
Adrian Frühwirth
26.1k75163
asked Sep 19 '13 at 10:52
JimJim
7,5012495179
1
AFAIK sed does not support the '+' quantifier. Instead you have to type the previous item twice: [0-9][0-9]* as anubhava does in his answer.
– Panos Rontogiannis
Sep 19 '13 at 11:50
1
possible duplicate of How to escape plus sign on mac os x (BSD) sed?
– Michael Foukarakis
Sep 19 '13 at 14:12
echo 'employee_id=1234' | cut -d '=' -f 2
– ALex_hha
Mar 11 '16 at 15:53
add a comment |
28
28
28
11
I want to extract the value pair from a key-value pair syntax but I can not.
Example I tried:
echo employee_id=1234 | sed 's/employee_id=([0-9]+)/1/g'
But this gives employee_id=1234 and not 1234 which is actually the capture group.
What am I doing wrong here? I also tried:
echo employee_id=1234| egrep -o employee_id=([0-9]+)
but no sucess.
regex linux sed grep
edited Sep 19 '13 at 18:41
Adrian Frühwirth
26.1k75163
asked Sep 19 '13 at 10:52
JimJim
7,5012495179
I want to extract the value pair from a key-value pair syntax but I can not.
Example I tried:
echo employee_id=1234 | sed 's/employee_id=([0-9]+)/1/g'
But this gives employee_id=1234 and not 1234 which is actually the capture group.
What am I doing wrong here? I also tried:
echo employee_id=1234| egrep -o employee_id=([0-9]+)
but no sucess.
regex linux sed grep
regex linux sed grep
edited Sep 19 '13 at 18:41
Adrian Frühwirth
26.1k75163
asked Sep 19 '13 at 10:52
JimJim
7,5012495179
edited Sep 19 '13 at 18:41
Adrian Frühwirth
26.1k75163
asked Sep 19 '13 at 10:52
JimJim
7,5012495179
edited Sep 19 '13 at 18:41
Adrian Frühwirth
26.1k75163
edited Sep 19 '13 at 18:41
Adrian Frühwirth
26.1k75163
edited Sep 19 '13 at 18:41
Adrian Frühwirth
26.1k75163
26.1k75163
asked Sep 19 '13 at 10:52
JimJim
7,5012495179
asked Sep 19 '13 at 10:52
JimJim
7,5012495179
asked Sep 19 '13 at 10:52
JimJim
7,5012495179
7,5012495179
1
AFAIK sed does not support the '+' quantifier. Instead you have to type the previous item twice: [0-9][0-9]* as anubhava does in his answer.
– Panos Rontogiannis
Sep 19 '13 at 11:50
1
possible duplicate of How to escape plus sign on mac os x (BSD) sed?
– Michael Foukarakis
Sep 19 '13 at 14:12
echo 'employee_id=1234' | cut -d '=' -f 2
– ALex_hha
Mar 11 '16 at 15:53
add a comment |
1
AFAIK sed does not support the '+' quantifier. Instead you have to type the previous item twice: [0-9][0-9]* as anubhava does in his answer.
– Panos Rontogiannis
Sep 19 '13 at 11:50
1
possible duplicate of How to escape plus sign on mac os x (BSD) sed?
– Michael Foukarakis
Sep 19 '13 at 14:12
echo 'employee_id=1234' | cut -d '=' -f 2
– ALex_hha
Mar 11 '16 at 15:53
1
1
AFAIK sed does not support the '+' quantifier. Instead you have to type the previous item twice: [0-9][0-9]* as anubhava does in his answer.
– Panos Rontogiannis
Sep 19 '13 at 11:50
AFAIK sed does not support the '+' quantifier. Instead you have to type the previous item twice: [0-9][0-9]* as anubhava does in his answer.
– Panos Rontogiannis
Sep 19 '13 at 11:50
1
1
possible duplicate of How to escape plus sign on mac os x (BSD) sed?
– Michael Foukarakis
Sep 19 '13 at 14:12
possible duplicate of How to escape plus sign on mac os x (BSD) sed?
– Michael Foukarakis
Sep 19 '13 at 14:12
echo 'employee_id=1234' | cut -d '=' -f 2
– ALex_hha
Mar 11 '16 at 15:53
echo 'employee_id=1234' | cut -d '=' -f 2
– ALex_hha
Mar 11 '16 at 15:53
add a comment |
5 Answers
5
active
oldest
votes
66
1. Use egrep -o:
echo 'employee_id=1234' | egrep -o '[0-9]+'
1234
2. using grep -oP (PCRE):
echo 'employee_id=1234' | grep -oP 'employee_id=K([0-9]+)'
1234
3. Using sed:
echo 'employee_id=1234' | sed 's/^.*employee_id=([0-9][0-9]*).*$/1/'
1234
answered Sep 19 '13 at 10:56
anubhavaanubhava
527k46325402
+1. But why does mine doesn't work. I used egrep -o
– Jim
Sep 19 '13 at 10:58
Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one afteremployee_id=
– Jim
Sep 19 '13 at 11:00
The last command in your comment does not work at all
– Jim
Sep 19 '13 at 11:01
1
None of your answers are relevant. The first I can not use because I need only the number afteremployee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use*in the digits part while I use+. Why does this matter?
– Jim
Sep 19 '13 at 11:06
1
2 worked for me as a way to get PHP version for use in an Ansible playbookphp -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"
– turrican_34
Dec 18 '18 at 14:04
|
show 5 more comments
18
To expand on anubhava's answer number 2, the general pattern to have grep return only the capture group is:
$ regex="$precedes_regexK($capture_regex)(?=$follows_regex)"
$ echo $some_string | grep -oP "$regex"
so
# matches and returns b
$ echo "abc" | grep -oP "aK(b)(?=c)"
b
# no match
$ echo "abc" | grep -oP "zK(b)(?=c)"
# no match
$ echo "abc" | grep -oP "aK(b)(?=d)"
answered Apr 6 '16 at 15:34
jayflojayflo
5401619
add a comment |
5
Using awk
echo 'employee_id=1234' | awk -F= '{print $2}'
1234
answered Sep 19 '13 at 11:29
JotneJotne
29.6k83245
add a comment |
3
You are specifically asking for sed, but in case you may use something else - any POSIX-compliant shell can do parameter expansion which doesn't require a fork/subshell:
foo='employee_id=1234'
var=${foo%%=*}
value=${foo#*=}
$ echo "var=${var} value=${value}"
var=employee_id value=1234
answered Sep 19 '13 at 14:00
Adrian FrühwirthAdrian Frühwirth
26.1k75163
add a comment |
1
use sed -E for extended regex
echo employee_id=1234 | sed -E 's/employee_id=([0-9]+)/1/g'
answered Nov 20 '18 at 5:34
commander Ghostcommander Ghost
111
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
66
1. Use egrep -o:
echo 'employee_id=1234' | egrep -o '[0-9]+'
1234
2. using grep -oP (PCRE):
echo 'employee_id=1234' | grep -oP 'employee_id=K([0-9]+)'
1234
3. Using sed:
echo 'employee_id=1234' | sed 's/^.*employee_id=([0-9][0-9]*).*$/1/'
1234
answered Sep 19 '13 at 10:56
anubhavaanubhava
527k46325402
+1. But why does mine doesn't work. I used egrep -o
– Jim
Sep 19 '13 at 10:58
Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one afteremployee_id=
– Jim
Sep 19 '13 at 11:00
The last command in your comment does not work at all
– Jim
Sep 19 '13 at 11:01
1
None of your answers are relevant. The first I can not use because I need only the number afteremployee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use*in the digits part while I use+. Why does this matter?
– Jim
Sep 19 '13 at 11:06
1
2 worked for me as a way to get PHP version for use in an Ansible playbookphp -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"
– turrican_34
Dec 18 '18 at 14:04
|
show 5 more comments
66
1. Use egrep -o:
echo 'employee_id=1234' | egrep -o '[0-9]+'
1234
2. using grep -oP (PCRE):
echo 'employee_id=1234' | grep -oP 'employee_id=K([0-9]+)'
1234
3. Using sed:
echo 'employee_id=1234' | sed 's/^.*employee_id=([0-9][0-9]*).*$/1/'
1234
answered Sep 19 '13 at 10:56
anubhavaanubhava
527k46325402
+1. But why does mine doesn't work. I used egrep -o
– Jim
Sep 19 '13 at 10:58
Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one afteremployee_id=
– Jim
Sep 19 '13 at 11:00
The last command in your comment does not work at all
– Jim
Sep 19 '13 at 11:01
1
None of your answers are relevant. The first I can not use because I need only the number afteremployee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use*in the digits part while I use+. Why does this matter?
– Jim
Sep 19 '13 at 11:06
1
2 worked for me as a way to get PHP version for use in an Ansible playbookphp -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"
– turrican_34
Dec 18 '18 at 14:04
|
show 5 more comments
66
66
66
1. Use egrep -o:
echo 'employee_id=1234' | egrep -o '[0-9]+'
1234
2. using grep -oP (PCRE):
echo 'employee_id=1234' | grep -oP 'employee_id=K([0-9]+)'
1234
3. Using sed:
echo 'employee_id=1234' | sed 's/^.*employee_id=([0-9][0-9]*).*$/1/'
1234
answered Sep 19 '13 at 10:56
anubhavaanubhava
527k46325402
1. Use egrep -o:
echo 'employee_id=1234' | egrep -o '[0-9]+'
1234
2. using grep -oP (PCRE):
echo 'employee_id=1234' | grep -oP 'employee_id=K([0-9]+)'
1234
3. Using sed:
echo 'employee_id=1234' | sed 's/^.*employee_id=([0-9][0-9]*).*$/1/'
1234
answered Sep 19 '13 at 10:56
anubhavaanubhava
527k46325402
edited Sep 19 '13 at 11:20
answered Sep 19 '13 at 10:56
anubhavaanubhava
527k46325402
answered Sep 19 '13 at 10:56
anubhavaanubhava
527k46325402
answered Sep 19 '13 at 10:56
anubhavaanubhava
527k46325402
527k46325402
+1. But why does mine doesn't work. I used egrep -o
– Jim
Sep 19 '13 at 10:58
Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one afteremployee_id=
– Jim
Sep 19 '13 at 11:00
The last command in your comment does not work at all
– Jim
Sep 19 '13 at 11:01
1
None of your answers are relevant. The first I can not use because I need only the number afteremployee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use*in the digits part while I use+. Why does this matter?
– Jim
Sep 19 '13 at 11:06
1
2 worked for me as a way to get PHP version for use in an Ansible playbookphp -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"
– turrican_34
Dec 18 '18 at 14:04
|
show 5 more comments
+1. But why does mine doesn't work. I used egrep -o
– Jim
Sep 19 '13 at 10:58
Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one afteremployee_id=
– Jim
Sep 19 '13 at 11:00
The last command in your comment does not work at all
– Jim
Sep 19 '13 at 11:01
1
None of your answers are relevant. The first I can not use because I need only the number afteremployee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use*in the digits part while I use+. Why does this matter?
– Jim
Sep 19 '13 at 11:06
1
2 worked for me as a way to get PHP version for use in an Ansible playbookphp -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"
– turrican_34
Dec 18 '18 at 14:04
+1. But why does mine doesn't work. I used egrep -o
– Jim
Sep 19 '13 at 10:58
+1. But why does mine doesn't work. I used egrep -o
– Jim
Sep 19 '13 at 10:58
Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one after
employee_id=– Jim
Sep 19 '13 at 11:00
Actually this will not work for me as in the same line I am regexing there is one more number. I only want the one after
employee_id=– Jim
Sep 19 '13 at 11:00
The last command in your comment does not work at all
– Jim
Sep 19 '13 at 11:01
The last command in your comment does not work at all
– Jim
Sep 19 '13 at 11:01
1
1
None of your answers are relevant. The first I can not use because I need only the number after
employee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use * in the digits part while I use +. Why does this matter?– Jim
Sep 19 '13 at 11:06
None of your answers are relevant. The first I can not use because I need only the number after
employee_id=, the second does not work at all and the third one picks the number and if I modify it, it gives what I need but the difference with mine is that you use * in the digits part while I use +. Why does this matter?– Jim
Sep 19 '13 at 11:06
1
1
2 worked for me as a way to get PHP version for use in an Ansible playbook
php -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"– turrican_34
Dec 18 '18 at 14:04
2 worked for me as a way to get PHP version for use in an Ansible playbook
php -v | grep -P -o "^PHPsK([0-9]{1}.?[0-9]{0,2}.?[0-9]{0,2})s"– turrican_34
Dec 18 '18 at 14:04
|
show 5 more comments
18
To expand on anubhava's answer number 2, the general pattern to have grep return only the capture group is:
$ regex="$precedes_regexK($capture_regex)(?=$follows_regex)"
$ echo $some_string | grep -oP "$regex"
so
# matches and returns b
$ echo "abc" | grep -oP "aK(b)(?=c)"
b
# no match
$ echo "abc" | grep -oP "zK(b)(?=c)"
# no match
$ echo "abc" | grep -oP "aK(b)(?=d)"
answered Apr 6 '16 at 15:34
jayflojayflo
5401619
add a comment |
18
To expand on anubhava's answer number 2, the general pattern to have grep return only the capture group is:
$ regex="$precedes_regexK($capture_regex)(?=$follows_regex)"
$ echo $some_string | grep -oP "$regex"
so
# matches and returns b
$ echo "abc" | grep -oP "aK(b)(?=c)"
b
# no match
$ echo "abc" | grep -oP "zK(b)(?=c)"
# no match
$ echo "abc" | grep -oP "aK(b)(?=d)"
answered Apr 6 '16 at 15:34
jayflojayflo
5401619
add a comment |
18
18
18
To expand on anubhava's answer number 2, the general pattern to have grep return only the capture group is:
$ regex="$precedes_regexK($capture_regex)(?=$follows_regex)"
$ echo $some_string | grep -oP "$regex"
so
# matches and returns b
$ echo "abc" | grep -oP "aK(b)(?=c)"
b
# no match
$ echo "abc" | grep -oP "zK(b)(?=c)"
# no match
$ echo "abc" | grep -oP "aK(b)(?=d)"
answered Apr 6 '16 at 15:34
jayflojayflo
5401619
To expand on anubhava's answer number 2, the general pattern to have grep return only the capture group is:
$ regex="$precedes_regexK($capture_regex)(?=$follows_regex)"
$ echo $some_string | grep -oP "$regex"
so
# matches and returns b
$ echo "abc" | grep -oP "aK(b)(?=c)"
b
# no match
$ echo "abc" | grep -oP "zK(b)(?=c)"
# no match
$ echo "abc" | grep -oP "aK(b)(?=d)"
answered Apr 6 '16 at 15:34
jayflojayflo
5401619
edited May 12 '16 at 22:25
answered Apr 6 '16 at 15:34
jayflojayflo
5401619
answered Apr 6 '16 at 15:34
jayflojayflo
5401619
answered Apr 6 '16 at 15:34
jayflojayflo
5401619
5401619
add a comment |
add a comment |
5
Using awk
echo 'employee_id=1234' | awk -F= '{print $2}'
1234
answered Sep 19 '13 at 11:29
JotneJotne
29.6k83245
add a comment |
5
Using awk
echo 'employee_id=1234' | awk -F= '{print $2}'
1234
answered Sep 19 '13 at 11:29
JotneJotne
29.6k83245
add a comment |
5
5
5
Using awk
echo 'employee_id=1234' | awk -F= '{print $2}'
1234
answered Sep 19 '13 at 11:29
JotneJotne
29.6k83245
Using awk
echo 'employee_id=1234' | awk -F= '{print $2}'
1234
answered Sep 19 '13 at 11:29
JotneJotne
29.6k83245
answered Sep 19 '13 at 11:29
JotneJotne
29.6k83245
answered Sep 19 '13 at 11:29
JotneJotne
29.6k83245
answered Sep 19 '13 at 11:29
JotneJotne
29.6k83245
29.6k83245
add a comment |
add a comment |
3
You are specifically asking for sed, but in case you may use something else - any POSIX-compliant shell can do parameter expansion which doesn't require a fork/subshell:
foo='employee_id=1234'
var=${foo%%=*}
value=${foo#*=}
$ echo "var=${var} value=${value}"
var=employee_id value=1234
answered Sep 19 '13 at 14:00
Adrian FrühwirthAdrian Frühwirth
26.1k75163
add a comment |
3
You are specifically asking for sed, but in case you may use something else - any POSIX-compliant shell can do parameter expansion which doesn't require a fork/subshell:
foo='employee_id=1234'
var=${foo%%=*}
value=${foo#*=}
$ echo "var=${var} value=${value}"
var=employee_id value=1234
answered Sep 19 '13 at 14:00
Adrian FrühwirthAdrian Frühwirth
26.1k75163
add a comment |
3
3
3
You are specifically asking for sed, but in case you may use something else - any POSIX-compliant shell can do parameter expansion which doesn't require a fork/subshell:
foo='employee_id=1234'
var=${foo%%=*}
value=${foo#*=}
$ echo "var=${var} value=${value}"
var=employee_id value=1234
answered Sep 19 '13 at 14:00
Adrian FrühwirthAdrian Frühwirth
26.1k75163
You are specifically asking for sed, but in case you may use something else - any POSIX-compliant shell can do parameter expansion which doesn't require a fork/subshell:
foo='employee_id=1234'
var=${foo%%=*}
value=${foo#*=}
$ echo "var=${var} value=${value}"
var=employee_id value=1234
answered Sep 19 '13 at 14:00
Adrian FrühwirthAdrian Frühwirth
26.1k75163
answered Sep 19 '13 at 14:00
Adrian FrühwirthAdrian Frühwirth
26.1k75163
answered Sep 19 '13 at 14:00
Adrian FrühwirthAdrian Frühwirth
26.1k75163
answered Sep 19 '13 at 14:00
Adrian FrühwirthAdrian Frühwirth
26.1k75163
26.1k75163
add a comment |
add a comment |
1
use sed -E for extended regex
echo employee_id=1234 | sed -E 's/employee_id=([0-9]+)/1/g'
answered Nov 20 '18 at 5:34
commander Ghostcommander Ghost
111
add a comment |
1
use sed -E for extended regex
echo employee_id=1234 | sed -E 's/employee_id=([0-9]+)/1/g'
answered Nov 20 '18 at 5:34
commander Ghostcommander Ghost
111
add a comment |
1
1
1
use sed -E for extended regex
echo employee_id=1234 | sed -E 's/employee_id=([0-9]+)/1/g'
answered Nov 20 '18 at 5:34
commander Ghostcommander Ghost
111
use sed -E for extended regex
echo employee_id=1234 | sed -E 's/employee_id=([0-9]+)/1/g'
answered Nov 20 '18 at 5:34
commander Ghostcommander Ghost
111
answered Nov 20 '18 at 5:34
commander Ghostcommander Ghost
111
answered Nov 20 '18 at 5:34
commander Ghostcommander Ghost
111
answered Nov 20 '18 at 5:34
commander Ghostcommander Ghost
111
111
add a comment |
add a comment |
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1
AFAIK sed does not support the '+' quantifier. Instead you have to type the previous item twice: [0-9][0-9]* as anubhava does in his answer.
– Panos Rontogiannis
Sep 19 '13 at 11:50
1
possible duplicate of How to escape plus sign on mac os x (BSD) sed?
– Michael Foukarakis
Sep 19 '13 at 14:12
echo 'employee_id=1234' | cut -d '=' -f 2
– ALex_hha
Mar 11 '16 at 15:53