[SOLVED] sed and lazy regex

CartmanYO · 10-06-2010, 05:55 AM

Hello! Here's my question. This command replaces "abb" substring

Code:

echo abbc | sed "s/.*b//g"

And this command doesn't work (lazy regex):

Code:

echo abbc | sed "s/.*?b//g"

Why? This one should replace "ab" substring.

druuna · 10-06-2010, 06:06 AM

Hi,

The ? needs to be escaped: echo abbc | sed "s/.*\?b//g"

Hope this helps.

CartmanYO · 10-06-2010, 06:17 AM

Thanks for your reply, druuna, but this fix doesn't help - the result of this command is still "c"

grail · 10-06-2010, 06:24 AM

What in your mind was the point of the question mark?

druuna · 10-06-2010, 06:26 AM

Hi,

This replaces a single character at the beginning of the line (^.) followed by a b with nothing. You'll end up with bc.
echo abbc | sed "s/^.b//"

I'm not sure if this is what your are looking for, but hope it helps.

CartmanYO · 10-06-2010, 06:30 AM

Sure, this will work in current case of "abbc". But what if the string is "123 abbc"? I'm interested in lazy regular expressions and i suppose they are bugged in sed.

druuna · 10-06-2010, 06:35 AM

Hi,

What are you trying to accomplish, maybe your approach isn't correct:

echo "1234 abbc" | sed "s/.*b/b/" will leave you with bc, regardless of what is in front of bc.

Hope this helps.

ghostdog74 · 10-06-2010, 06:36 AM

sed does not support non-greediness. To do that, you use [^]

Code:

# echo abbc | sed 's/.*[^b]b//g'
bc

CartmanYO · 10-06-2010, 06:41 AM

OK, thanks. That was all i need to know.

grail · 10-06-2010, 06:46 AM

Are you perhaps trying to catch everything up until the first b and and then also the first b?
try:

Code:

echo "123 abbc" | sed 's/[^b]*b//'