I wrote a regular exprssion:
(?<value>(?<=start=)\d+)
I'tested it in Exprsso and it works When Itried to use it in BASH script, there are no result, althogh the data is the same as in the test with Expresso. Can a gruoping be used with grep, and return array like result?
grep -Po "(?<value>(?<=start=)\d+)" page.txt > result.txt
Is this the way it is used?
Edit - example data (working)
www.example.com/check?var=test&start=11
If you're looking to get the numbers following "start=", you don't need a named capture group. You don't even need capturing parentheses, that's really what -o does.
Simplify:
grep -Po '(?<=start=)\d+'
More examples would be:
echo 'some text here:11' | grep -Po '(?<=here:)\d+'
11
echo 'some text here:apple' | grep -Po '(?<=here:)\w+'
apple
The Named Capturing Group is not necessary here. Using the -o option shows only the matching part that matches the pattern, so with the combination of the Positive Lookbehind assertion and the -o command line option, you do not need a capturing group. The lookbehind does not consume characters in the string, but only asserts whether a match is possible or not.
echo 'foo.com/check?var=test&start=11' | grep -Po '(?<=start=)\d+'
# 11
Alternatively, \K resets the starting point of the reported match and any previously consumed characters are no longer included. e.g. throws away everything that it has matched up to that point.
grep -Po 'start=\K\d+'
Another way of doing positive lookbehind with \K:
echo 'foo.com/check?var=test&start=11' | grep -Po 'start=\K\d+'
11
