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sed help, Find a pattern, replace it with same text minus leading 0

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# 1  
Old 07-10-2012
Question sed help, Find a pattern, replace it with same text minus leading 0
HI Folks,

I'm looking for a solution for this issue.

I want to find the Pattern 0[0-9]/ and replace it with [0-9]/. I'm just removing the leading zero. I can find the Pattern but it always puts literal value as a replacement.

What am I missing??


Code:
sed -e s/0[0-9]\//[0-9]\//g File1 > File2

Moderator's Comments:
Mod Comment edit by bakunin: Please view this code tag video for how to use code tags when posting code and data.
Last edited by bakunin; 07-10-2012 at 11:59 AM..
# 2  
Old 07-10-2012
sed doesn't work that way. putting [0-9] in the output pattern will not cause it to substitute any of the input.

To do that, you want a backreference.

Code:
sed 's#0\([0-9]\)/#\1/#'

This User Gave Thanks to Corona688 For This Post:
SirHenry1
# 3  
Old 07-10-2012
You need the "\(...\)" device for this. This way you can replicate parts of the matched regexp in the substitution:

Code:
sed 's/0\([0-9]\)/\1/'

will replace any "0" followed by a digit with this digit and omit the leading "0".

I hope this helps.

bakunin
This User Gave Thanks to bakunin For This Post:
SirHenry1
# 4  
Old 07-10-2012
Thanks, It worked.
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