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Regular expression for 6 digit number present in a line


 
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# 1  
Old 08-29-2013
Regular expression for 6 digit number present in a line

Hello Team,
i have a file test1.txt, in which i have to grep only the 6 digit number from it,
Could you pls help in this.

Code:
$cat test1.txt
  <description>R_XYZ_1.6 r370956</description>
$ grep "[0-9]\{6\}" test1.txt
  <description>R_XYZ_1.6 r370956</description>

i need output as 370956.

Regards,
Chandana
# 2  
Old 08-29-2013
if you're using gnu grep, grep -o returns the matched text.
if you don't have gnu grep available try, perl -ne 'print "$1\n" if /(\d{6})/' test1.txt
# 3  
Old 08-29-2013
Thanks,

Got it.

Code:
$ cat test1.txt |sed -n -e 's/.*\([0-9]\{6\}\).*/\1/p'
370956

# 4  
Old 08-29-2013
using awk
Code:
awk '{match($0,"[0-9][0-9][0-9][0-9][0-9][0-9]",a);print a[0]}' test1.txt
370956

PS do not cat the file to sed/awk, add file behind, like my example.

Last edited by vbe; 08-29-2013 at 06:18 AM.. Reason: typo hehe
This User Gave Thanks to Jotne For This Post:

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