grep: group capturing
RegexLinuxBashGrepRegex Problem Overview
I have following string:
{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}
and I need to get value of "scheme version", which is 1234 in this example.
I have tried
grep -Eo "\"scheme_version\":(\w*)"
however it returns
"scheme_version":1234
How can I make it? I know I can add sed call, but I would prefer to do it with single grep.
Regex Solutions
Solution 1 - Regex
You'll need to use a look behind assertion so that it isn't included in the match:
grep -Po '(?<=scheme_version":)[0-9]+'
Solution 2 - Regex
This might work for you:
echo '{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}' |
sed -n 's/.*"scheme_version":\([^}]*\)}/\1/p'
1234
Sorry it's not grep, so disregard this solution if you like.
Or stick with grep and add:
grep -Eo "\"scheme_version\":(\w*)"| cut -d: -f2
Solution 3 - Regex
I would recommend that you use jq for the job. jq is a command-line JSON processor.
$ cat tmp
{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}
$ cat tmp | jq .scheme_version
1234
Solution 4 - Regex
As an alternative to the positive lookbehind method suggested by SiegeX, you can reset the match starting point to directly after scheme_version": with the \K escape sequence. E.g.,
$ grep -Po 'scheme_version":\K[0-9]+'
This restarts the matching process after having matched scheme_version":, and tends to have far better performance than the positive lookbehind. Comparing the two on regexp101 demonstrates that the reset match start method takes 37 steps and 1ms, while the positive lookbehind method takes 194 steps and 21ms.
You can compare the performance yourself on regex101 and you can read more about resetting the match starting point in the PCRE documentation.
Solution 5 - Regex
To avoid using greps PCRE feature which is available in GNU grep, but not in BSD version, another method is to use ripgrep, e.g.
$ rg -o 'scheme_version.?:(\d+)' -r '$1' <file.json
1234
> -r Capture group indices (e.g., $5) and names (e.g., $foo).
Another example with Python and json.tool module which can validate and pretty-print:
$ python -mjson.tool file.json | rg -o 'scheme_version[^\d]+(\d+)' -r '$1'
1234
Related: Can grep output only specified groupings that match?
Solution 6 - Regex
You can do this:
$ echo '{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}' | awk -F ':' '{print $4}' | tr -d '}'
Solution 7 - Regex
Improving @potong's answer that works only to get "scheme_version", you can use this expression :
$ echo '{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}' | sed -n 's/.*"_id":["]*\([^(",})]*\)[",}].*/\1/p'
scheme_version
$ echo '{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}' | sed -n 's/.*"_rev":["]*\([^(",})]*\)[",}].*/\1/p'
4-cad1842a7646b4497066e09c3788e724
$ echo '{"_id":"scheme_version","_rev":"4-cad1842a7646b4497066e09c3788e724","scheme_version":1234}' | sed -n 's/.*"scheme_version":["]*\([^(",})]*\)[",}].*/\1/p'
1234