My data contains lines like this:
55 511 00,"805, 809, 810, 839, 840",J223,201,338,116,16,200,115,6,P,S,"8,5","25,74",47,242,"55,7"
I have tried ,"(.*)",
as a regular expression, but it captures too much of the line. This expression currently returns:
,"805, 809, 810, 839, 840",J223,201,338,116,16,200,115,6,P,S,"8,5","25,74",
but what I really want is just the first quoted string. Valid results would be:
,"805, 809, 810, 839, 840",
805, 809, 810, 839, 840
How can I capture only that first match?
Try "([^"]+)
. Group 1 will match 805, 809, 810, 839, 840
/"([^"]+)"/
Will do the job! Everything between the "-s
Your regex is greedy, the .* will get everything up until the final "
So to make it non-greedy, add a ? at the end of the bracketed part:
,"(.*?)",
Which should stop it as soon as it reaches the next "
There are many ways to handle this, but the simplest and most generic is to use a non-greedy match if your regular expression engine supports it. If it does not, you have to build an expression that knows a lot more about the structure of your data.
Here's an example using Perl-compatible regular expressions to split the output:
$ pcregrep -o '"(.*?)"' /tmp/foo | head -n1
"805, 809, 810, 839, 840"
Here's another example that uses pure Perl:
$ perl -ne 'print "$1\n" if /(".*?")/' /tmp/foo
"805, 809, 810, 839, 840"
Here's a third example that uses POSIX extended regular expressions, but which does not support non-greedy matches.
$ egrep -o '("[^"]+")' /tmp/foo | head -n1
"805, 809, 810, 839, 840"
You may also want to consider splitting your input into fields, and then testing each field until you find a match. A lot just depends on what facilities you have at your disposal.
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