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 success.
1. Use grep -Eo: (as egrep is deprecated)
echo 'employee_id=1234' | grep -Eo '[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
To expand on anubhava's answer number 2, the general pattern to have grep return only the capture group is:
$ regex="$precedes_regex\K($capture_regex)(?=$follows_regex)"
$ echo $some_string | grep -oP "$regex"
so
# matches and returns b
$ echo "abc" | grep -oP "a\K(b)(?=c)"
b
# no match
$ echo "abc" | grep -oP "z\K(b)(?=c)"
# no match
$ echo "abc" | grep -oP "a\K(b)(?=d)"
Using awk
echo 'employee_id=1234' | awk -F= '{print $2}'
1234
use sed -E for extended regex
echo employee_id=1234 | sed -E 's/employee_id=([0-9]+)/\1/g'
You are specifically asking for sed, but in case you may use something else - any POSIX-compliant shell can do parameter expansion which doesn't require a fork/subshell:
foo='employee_id=1234'
var=${foo%%=*}
value=${foo#*=}
$ echo "var=${var} value=${value}"
var=employee_id value=1234
Related
How do I grep strings in between nested brackets using bash? Is it possible without the use of loops? For example, if I have a string like:
[[TargetString1:SomethingIDontWantAfterColon[[TargetString2]]]]
I wish to grep only the two target strings inside the [[]]:
TargetString1
TargetString2
I tried the following command which cannot get TargetString2
grep -o -P '(?<=\[\[).*(?=\]\])'|cut -d ':' -f1
With GNU's grep P option:
grep -oP "(?<=\[\[)[\w\s]+"
The regex will match a sequence of word characters (\w+) when followed by two brackets ([[). This works for your sample string, but will not work for more complicated constructs like:
[[[[TargetString1]]TargetString2:SomethingIDontWantAfterColon[[TargetString3]]]]
where only TargetString1 and TargetString3 are matched.
To extract from nested [[]] brackets, you can use sed
#!/bin/bash
str="[[TargetString1:SomethingIDontWantAfterColon[[TargetString2]]]]"
echo $str | grep -o -P '(?<=\[\[).*(?=\]\])'|cut -d ':' -f1
echo $str | sed 's/.*\[\([^]]*\)\].*/\1/g' #which works only if string exsit between []
Output:
TargetString1
TargetString2
You can use grep regex grep -Eo '\[\[\w+' | sed 's/\[\[//g' for doing this
[root#localhost ~]# echo "[[TargetString1:SomethingIDontWantAfterColon[[TargetString2]]]]" | grep -Eo '\[\[\w+' | sed 's/\[\[//g'
TargetString1
TargetString2
[root#localhost ~]#
I want to find the set of words that contain two consecutive e’s, and also contains two y’s.
So far i got to /eeyy/
Alteration with ERE:
$ echo evyyree | grep -E '.*ee.*yy|.*yy.*ee'
evyyree
$ echo eveeryy | grep -E '.*ee.*yy|.*yy.*ee'
eveeryy
If the match needs to be in the same word, you can do:
$ echo "eee yyyy" | grep -E 'ee[^[:space:]]*yy|yy[^[:space:]]*ee' # no match
$ echo "eeeyyyy" | grep -E 'ee[^[:space:]]*yy|yy[^[:space:]]*ee'
eeeyyyy
Then only that word:
$ echo 'eeeyy heelo' | grep -Eo 'ee[^[:space:]]*yy|yy[^[:space:]]*ee'
eeeyy
Pipe it:
$ echo eennmmyy | grep ee | grep yy
eennmmyy
awk approach to match all words that contain both ee and yy:
s="eennmmyy heello thees-whyy someyy"
echo $s | awk '{for(i=1;i<=NF;i++) if($i~/ee/ && $i~/yy/) print $i}'
The output:
eennmmyy
thees-whyy
The only sensible and extensible way to do this is with awk:
awk '/ee/&&/yy/' file
Imagine trying to do it the grep way if you also had to find zz. Here's awk:
awk '/ee/&&/yy/&&/zz/' file
and here's grep:
grep -E 'ee.*yy.*zz|ee.*zz.*yy|yy.*ee.*zz|yy.*zz.*ee|zz.*yy.*ee|zz.*ee.*yy' file
Now add a 4th additional string to search for and see what that looks like!
I have an input string in the following format:
bugfix/ABC-12345-1-00
I want to extract "ABC-12345". Regex for that format in C# looks like this:
.\*\\/([A-Z]+-[0-9]+).\*
How can I do that in a bash script? I've tried sed and awk but had no success because I need to extract value from the capturing group and skip the rest.
If your grep supports -P then you could use the below grep commands.
$ echo 'bugfix/ABC-12345-1-00' | grep -oP '/\K[A-Z]+-\d+'
ABC-12345
\K keeps the text matched so far out of the overall regex match.
$ echo 'bugfix/ABC-12345-1-00' | grep -oP '(?<=/)[A-Z]+-\d+'
ABC-12345
(?<=/) Positive lookbehind which asserts that the match must be preceded by a / symbol.
Through sed,
$ echo 'bugfix/ABC-12345-1-00' | sed 's~.*/\([A-Z]\+-[0-9]\+\).*~\1~'
ABC-12345
echo "bugfix/ABC-12345-1-00"| perl -ane '/.*?([A-Z]+\-[0-9]+).*/;print $1."\n"'
You could try something like:
echo "bugfix/ABC-12345-1-00" | egrep -o '[A-Z]+-[0-9]+'
OUTPUT:
ABC-12345
If you do not like to use regex, you can use this awk:
echo "bugfix/ABC-12345-1-00" | awk -F\/ '{print $NF}'
ABC-12345-1-00
Or just this:
awk -F\/ '$0=$NF'
I wanted to extract 12 from a text like "abc_12_1". I am trying like this
echo "abc_12_1" | grep -Eo '[a-zA-Z]+_[0-9]+_1'
abc_12_1
But I am not able to select the digit after first _ in string, the output of above command is whole string. I am looking for some alternative in grep which I have in following Perl pattern matching.
perl -e '"abc_55_1" =~ m/[a-zA-Z]+_([0-9]+)_1/ ; print $1'
55
Is it possible with grep?
Using perl:
$ echo "abc_12_1" | perl -lne 'print /_(\d+)_/'
12
or grep:
$ echo "abc_12_1" | grep -oP '(?<=_)\d+(?=_)'
12
You could use cut:
cut -d_ -f2 <<< "abc_12_1"
Using grep:
grep -oP '(?<=_).*?(?=_)' <<< "abc_12_1"
Both would yield 12.
One way is to use awk
echo "abc_12_1" | awk -F_ '{print $2}'
12
Or grep
echo "abc_12_1" | grep -o "[0-9][0-9]"
12
Using grep with extended regex
grep -oE "[0-9]{2}" # Get only hits with two digits
grep -oE "[0-9]{2,}" # Get hits with two or more digits
I make a cat of a file and apply on it a grep with a regular expression like this
cat /tmp/tmp_file | grep "toto.titi\[[0-9]\+\].tata=55"
the command display the following output
toto.titi[12].tata=55
is it possible to modify my grep command in order to extract the number 12 as displayed output of the command?
You can grab this in pure BASH using its regex capabilities:
s='toto.titi[12].tata=55'
[[ "$s" =~ ^toto.titi\[([0-9]+)\]\.tata=[0-9]+$ ]] && echo "${BASH_REMATCH[1]}"
12
You can also use sed:
sed 's/toto.titi\[\([0-9]*\)\].tata=55/\1/' <<< "$s"
12
OR using awk:
awk -F '[\\[\\]]' '{print $2}' <<<"$s"
12
use lookahead
echo toto.titi[12].tata=55|grep -oP '(?<=\[)\d+'
12
without perl regex,use sed to replace "["
echo toto.titi[12].tata=55|grep -o "\[[0-9]\+"|sed 's/\[//g'
12
Pipe it to sed and use a back reference:
cat /tmp/tmp_file | grep "toto.titi\[[0-9]\+\].tata=55" | sed 's/.*\[(\d*)\].*/\1/'