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Shell Programming and Scripting

How to replace multiple numbers?

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# 8  
Old 12-31-2010
Something like this,

Code:
 perl -nle '$v="00";s/(ASX1[89]\d{6})(07|12)(\d{3})/$1$v$3/g;print $_;' infile

Hi Experts,

Is there any other way to specify number after $1 or \1 ?
# 9  
Old 12-31-2010
Use curly brackets?
Code:
perl -pe 's/^(ASX1[89]\d{6})(07|12)/${1}00/' infile

or
Code:
perl -pe 's/(?<=^ASX1[89]\d{6})(07|12)/00/' infile

This User Gave Thanks to Scrutinizer For This Post:
pravin27
# 10  
Old 01-01-2011
Quote:
Originally Posted by anurag.singh
Code:
sed 's/^\(ASX1[89]......\)07\(.*\)/\100\2/;s/^\(ASX1[89]......\)12\(.*\)/\100\2/' inputFile

Quote:
Originally Posted by Scrutinizer
You can leave out \2:
Code:
sed 's/^\(ASX1[89]......\)07/\100/;s/^\(ASX1[89]......\)12/\100/' infile

I tried the above solutions and Smilie

Is it possible to make it one instruction instead of two? i mean in the first instruction can i make it check for 07 or 12, something like:
Code:
sed 's/^\(ASX1[89]......\)[07|12]/\100/' infile

i tried it and didn't work Smilie
# 11  
Old 01-01-2011
[07|12] means a single character, being either 0,7,|,1, or 2. So that would not work. There are other examples in this thread, among others with GNU sed, but imho it cannot be done in standard sed..
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