sed / regex question

whysyn · 06-28-2005, 10:23 AM

hello,

i can usually hack my way through regular expressions, but this is troubling me.

here is a sample of text i'm working with:

Code:

#testsed.txt
time:4;
time:37;
time:9;
time:108;

this regex does not work to replace the number:
sed -e 's/time:[0-9]+;/time:99;/' testsed.txt

but this does:
sed -e 's/time:[0-9]*;/time:99;/' testsed.txt

why? as far as i knew, '+' means match 1 or more times. '*' means match any number of times, including none. wouldn't using a '*' cause an empty time field to be filled in with the replacement value? i wouldn't want that to happen...

thanks!

Berhanie · 06-28-2005, 11:09 AM

Add a -r flag since you're using extended regexps:

Code:

 sed -r -e 's/time:[0-9]+;/time:99;/' testsed.txt

or, if you want to use basic regexps, you need to escape the +:

Code:

 sed -e 's/time:[0-9]\+;/time:99;/' testsed.txt

ilikejam · 06-28-2005, 11:19 AM

Hi.

Looking at the man page for grep, which has a bit on regular expressions in general, it seems the + character has to be escaped to use it in that context (but * doesn't). It probably works the same for sed.

Try:
sed -e 's/time:[0-9]\+;/time:99;/' testsed.txt

Dave

whysyn · 06-28-2005, 02:11 PM

works like a charm. thanks, guys!