Regular exp in awk

Shell Programming and Scripting
1

Hi all,
One small doubt, reg exp in awk, is it possible to extract the match of regular expression like in perl?

eg:
B R16288 Map3_NoCat
B R16067 Map4_NoCat
B R16647 Map3_NoCat
B R16450 Map3_NoCat
B R16106 Map6_NoCat
B R16000 Map3_NoCat
B R16395 Map3_NoCat
B R16243 Map3_NoCat
B R16023 Map12_NoCat
B R16421 Map3_NoCat

if I wanna extract the digits after Map, in perl I could say

 /Map(\d*)/  print $1

is there any direct way in awk for it?

2

awk:

awk -F"[ _]" '{sub(/Map/,"",$3); print $3}' infile

sed:

sed 's/.*Map\([0-9]\+\)_.*/\1/g' infile
3 
#!/bin/bash
while read -r line
do
   line=${line##*Map}
   echo ${line%_*}
done < myfile
4

@zaxxon:
in your awk script

Code:
  awk -F"[ _]" '{sub(/Map/,"",$3); print $3}' infile

you r deleting the 'map', but what i want is to store the digits after map in order to do some logical test (say if the digit is > 5 then print the whole line)

Geo

5

No, in traditional awk you can't do that, as awk doesn't support back references. If you use Gnu awk (gawk), you could use the gensub()-function, which supports back references. If you don't have gawk, use sed instead.

HTH

Chris

6

@gvj

Yes, since that was what you asked for - if you have additional requests, please let it not sound as if there was missing someting in the answer to the original question, ty :wink:

awk -F"[ _]" '{a=$0; sub(/Map/,"",$3); if($3 > 5) print a}' infile
B R16106 Map6_NoCat
B R16023 Map12_NoCat
7

Another approach:

awk -F "Map|_" '$2 > 5' file