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Subject: | RE: grep -P regexp problem |
Date: | Wed, 4 Mar 2009 13:54:29 -0000 |
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Andriy Sen wrote: > Below is an example of the problem. > > G:\>cat test.s > a > 1 >=20 > G:\>cat test.s | grep -P "[^0]1" > a > 1 This is not cygwin-specific, so it is really OT for this list, that being said... grep -P treats the whole input as a single string, and outputs the line (or lines) containing the match for the pattern. [^0] matches=20 ANYTHING except 0, including linefeeds. In your case, the [^0] is matching the linefeed preceding the 1. That linefeed is considered part of the line "a\n", so that line is included in the output. In other words, although it looks like there are two matches output, in fact there is only one match, and that is "a\n1\n" Assuming you wish to match single lines containing a character other=20 than 0 followed by a 1, you probably want the pattern to be '[^0\n]1' It's probably a bit clearer if the test file is a bit bigger: $ echo -e 'a\n1\n2\n3\n4\n1\n2\n21\n' > test.txt $ grep -P '[^0]1' test.txt=20 a 1 4 1 21 This output contains 3 matches "a\n1\n" "4\n1\n" and "21\n", whereas: $ grep -P '[^0\n]1' test.txt=20 21 only matches single lines with a 1 that follows anything but 0. Phil --=20 This email has been scanned by Ascribe PLC using Microsoft Antigen for Exch= ange. -- Unsubscribe info: http://cygwin.com/ml/#unsubscribe-simple Problem reports: http://cygwin.com/problems.html Documentation: http://cygwin.com/docs.html FAQ: http://cygwin.com/faq/
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