How to print arguments along with spaces using awk

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# 8  
Old 01-17-2008
Originally Posted by Klashxx
jisha you need to be more polite , Yogesh was trying to help u.

First , your regex in wrong it'll macths all lines that contains the "aaa" not only first column,...
anyway many ways to do , one could be:
awk '/^aaa/{sub(/^aaa */,"",$0);print}' file

Thank you Klashxx

but code doesnot print the spaces in the begining .. Also when i ran this in command promt other than the starting spaces everything else is working .. But am supposed to run this in a script where the pattern is received from a function call .. When i executed the script, i got error saying unamatched ..

The fields in the file are unique ..Nothing is repeated ..Also the pattern for matching is always at teh first of coloumn in my file ..

Once again thanks for the effort ...
# 9  
Old 01-17-2008
Originally Posted by radoulov
awk 'sub(/^aaa/,"")' filename

I'm not sure if you want the space after the pattern (if any),
in that case:

awk 'sub(/^aaa */,"")' filename

HI radoulov,

Thank you for the reply ..
Finally i have made it as given below :

awk 'sub(/^('"${name}"')/,"")'

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