Perl regexp replace?

Question

I'm checking to see if I can replace an include like this:

#include <pathFoo/whatever.H>

with:

#include "whatever.H"

to do so I would use the -i switch, but to check I am doing it correctly I am simply using the -p switch without -i. I have the following command:

perl -p -e 's/<pathFoo\/\(.*\)>/"$1"' thefile

but this isn't quite working and i'm not exactly sure which part is off?

ErikR · Accepted Answer · 2012-12-05 00:02:52Z

4

You don't want to escape the parens, i.e.:

perl -p -e 's/<pathFoo\/(.*)>/"$1"/' thefile

should work for you.

Also note the ending /.

answered Dec 5, 2012 at 0:02

ErikR

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ikegami · Accepted Answer · 2012-12-05 00:11:12Z

1

perl -i~ -pe's!<pathFoo/(.*)>!"$1"!' file

The following is safer:

perl -i~ -pe's!#include\s+\K<pathFoo/(.*)>!"$1"!' file

edited Dec 5, 2012 at 0:11

answered Dec 5, 2012 at 0:05

ikegami

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TLP · Accepted Answer · 2012-12-05 00:18:40Z

1

Since TIMTOWTDI, here's a double substitution. Or rather a substitution and a transliteration:

perl -pe 's|^#include\s+<\KpathFoo/|| && tr/<>/""/'

So, just remove the pathFoo/ part first, and if that succeeds, then transliterate the <> characters to quotes.

answered Dec 5, 2012 at 0:18

TLP

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ddoxey · Accepted Answer · 2012-12-05 00:09:38Z

0

I think this will get you there:

perl -pi -e 's{#include[ ]+<.+?/([^/]+)>}{#include "$1"}g' file

answered Dec 5, 2012 at 0:09

ddoxey

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