It may be duplicate question, as I saw many with same title, but I cant find the error in my script by those questions...

My file is

for (( c=15; c<=60; c   ))
do 
while read -r i
do
   echo $i | awk -v var="$c" -F, 'var ~ /@/ {print $0}' >> $c.csv
done < asia.csv
done

when I run like sh -x clean.sh

it show

  read -r i
  awk -v var=15 -F, 'var ~ /@/ {print $0}'
  echo bellerock-casino.asia,CSC Corporate Domains R25-ASIA '(299),[email protected],,DNS1.CURACAOHOSTING.COM|DNS.CURACAOHOSTING.COM|DNS.CURACAOHOSTING.NET|,,20-Jan-2014' 15:34:56 UTC,01-Mar-2015 20:05:35 UTC,,2014-01-20 15:34:56 UTC,2015-03-01 20:05:35 UTC,CLIENT TRANSFER PROHIBITED,2014-04-17 07:00:00 UTC,Registrant Name:Limited 'Cassini|Registrant' Organization:Cassini 'Limited|Registrant' Address:PO Box 606 Mezzanine 'West|Registrant' 'City:Gibraltar|Registrant' 'Country/Economy:GB|Registrant' Postal 'Code:1000|Registrant' 'Phone: 44.7624496408|Registrant' 'FAX: 44.1481822895|Registrant' E-mail:[email protected],[email protected],Limited Cassini,Cassini Limited,PO Box 606 Mezzanine West,,,,Gibraltar,,1000,UNITED KINGDOM,441481822895,,447624496408,,Administrative Name:Toohey 'Michael|Administrative' Organization:Australasian Gaming Specialists Pty 'Limited|Administrative' Address:3 Kimberley 

for some reasons inside the ' ' var won't print 15

Any idea why so...

i.e. my script is checking if 15th field (and so on 16,17,18th) has @ or not... if it is the whole result should print in 15.csv

which is empty at the moment...

CodePudding user response：

If you want to check the 15th field instead of 15 itself, you should use $var, not just var.

awk -v var="$c" -F, '$var ~ /@/ {print $0}' <<<"$i" >>"$c.csv"

Consider shortening your loop to run awk only once per value of c, instead of once per line per value of c:

#!/usr/bin/env bash
for (( c=15; c<=60; c   )); do 
  awk -v var="$c" -F, '$var ~ /@/ {print $0}' >"$c.csv" <asia.csv
done