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matching two consecutive identical characters in awk
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Join Date: Oct 2008
Posts: 52
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Junior Poster in Training
i have come across a question which asks to write an awk command that displays all records having 2nd and 3rd characters same.
i could write the solution in sed, using tagged regular expression, as follows :
However, as far as i know, tagged expression are not applicable in awk. I wrote the solution in awk using substr() :
What i want to know is is there any other way in awk to solve this question? I just want to use a regular expression as i did in sed.
Also, is there any other way to write this in sed as well?
Thank you
i could write the solution in sed, using tagged regular expression, as follows :
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sed -n '/^.\(.\)\1.*$/' emp.list
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awk '{ if (substr($0,2,1) == substr($0,3,1)) print $0; }' emp.list
Also, is there any other way to write this in sed as well?
Thank you
Last edited by Bhoot; Mar 23rd, 2009 at 12:44 am. Reason: I forgot to quote the code.
Bhoot
You seem to be right; awk/gawk do seem to have limited REs.
But I do have a wiseacre solution.
But I do have a wiseacre solution.
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Join Date: Oct 2008
Posts: 52
Reputation:
Solved Threads: 1
Junior Poster in Training
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Originally Posted by Fest3er 
You seem to be right; awk/gawk do seem to have limited REs.
But I do have a wiseacre solution.
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lol..i couldnt help myself from laughing when i saw your solution
a good one; but certainly not applicable for me
anyways thanks for that too
Bhoot
>I wrote the solution in awk using substr() :
A slightly different approach. Since the targets to compare are found always in the beginning, use only the first field. Separate individual characters using split and then compare.
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awk '{ if (substr($0,2,1) == substr($0,3,1)) print $0; }' emp.list
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regular expressions are usually not needed for string manipulations.
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