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regular expression grepping lines with VARIOUS number of blanks


 
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# 8  
Old 08-11-2009
@Franklin52
What I posted works on my GNU sed on my Debian box. For AIX I had to remove the backslash in front of the +.

Code:
echo "*        Runjob=1" |grep -E "^\*[[:space:]]+Runjob=1"

Which works on the GNU sed too I just saw. I think without backslash is correct anyway since it should be interpreted and not be protected.
# 9  
Old 08-11-2009
Hi,

I am sorry to say, but it none of these commands work.
Maybe the reason is because I need it in bash-shell?!

For instance when I try following code

Code:
egrep '^*[ ]+Runjob=1' test.ini

or
Code:
egrep '*[ ]+Runjob=1' test.ini

I get the return message

Code:
*?+ not preceded by valid expression

Can someone help me?

---------- Post updated at 05:31 AM ---------- Previous update was at 05:28 AM ----------

following code worked:

Code:
 
grep -e "\*[ ]+*.*Runjob=1"

I'll try it right now, again thanks a lot to all of you Smilie

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