GnuWin32 sed 4.1.4 regexp matching - UNIX for Dummies Questions & Answers

Simerian · #1 (**permalink**) 06-22-2005

I am using GnuWin32 sed and am having trouble with the regexp - i.e., they don't behave the same way as in UNIX (POSIX and and all that). I have a stream of data, e.g.:

11111'222?'22'33?'333'44444'55555'

I want to insert a \n after those apostrophes that are *not* preceded by a ?.

Expected output:

11111'
222?'22'
33?'333'
44444'
55555'

I used the substitution:

s/\([^?]\)'/\1'\n/g

this should be OK but it didn't work... it replaced those that *did* have a ? in front - I want the opposite.

Any gurus with a suggestion? (apart from sticking to UNIX)

reborg · #2 (**permalink**) 06-22-2005

I can't imagine that that code did what you want in Unix either.

You need to escape your single quotes for them to be recognised as characters in the file and not as quotes around part of the command.

Ygor · #3 (**permalink**) 06-22-2005

Seems to work fine...

Code:

$ sed --version
GNU sed version 4.1.4
Copyright (C) 2003 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE,
to the extent permitted by law.

$ head file1 sed1
==> file1 <==
11111'222?'22'33?'333'44444'55555'

==> sed1 <==
s/\([^?]\)'/\1'\n/g

$ sed -f sed1 file1
11111'
222?'22'
33?'333'
44444'
55555'

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