problem with sed

mierdatuti · #1 (**permalink**) 01-22-2009

Hi,

I have this:

bash-2.03$ echo $kk
<programme start="20090124053000 +0100" channel=\CTele-205.laguiatv.com">

I would like to extract= 20090124053000

But I do:

echo $kk |sed -e 's/.*start="$.*$ .*/\1/'

and it gives me:

20090124053000 +0100"

I can't understand why this command doesn't works. Any help??

Many thanks!

zaxxon · #2 (**permalink**) 01-22-2009

It does not work because sed takes your 1st grouping between the ( and ) and finds a .* which means any number of any characters and even a blank. Then you try to tell it that a blank shows up, marking the end for the grouping, which it can't parse correctly since there could be a blank inside the brackets too (at least I explain it myself with this thesis of ambigous input). So you have to tell it, that the pattern/grouping exists of any character but NOT a blank:

Code:

..| sed -e 's/.*start="\([^ ]*\) .*/\1/'

mierdatuti · #3 (**permalink**) 01-22-2009

Quote:

Originally Posted by zaxxon

It does not work because sed takes your 1st grouping between the ( and ) and finds a .* which means any number of any characters and even a blank. Then you try to tell it that a blank shows up, marking the end for the grouping, which it can't parse correctly since there could be a blank inside the brackets too (at least I explain it myself with this thesis of ambigous input). So you have to tell it, that the pattern/grouping exists of any character but NOT a blank:

Code:

..| sed -e 's/.*start="\([^ ]*\) .*/\1/'

I changed the . in front of the * inside the $ and $ to [^ ] which means "anything but no blank". This should work:

Code:

echo ${kk}| sed -e 's/.*start="\([^ ]*\) .*/\1/'
20090124053000

Many thanks! It works! and than you for the explication

JerryHone · #4 (**permalink**) 01-22-2009

It's known as "greedy matching". The regular expression will always try and match the biggest string that satisfies the specified pattern, which in your case includes the embedded space.
In Perl, you can specify "non greedy" matching by following the * with a ?.

I think you could also have done...

Code:

echo ${kk}| sed -e 's/.*="\([0-9]*\) .* /\1/'

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