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How to filter out consecutive occurence of digits from file?

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# 1  
Old 07-13-2016
How to filter out consecutive occurence of digits from file?
Hello,

I have a text file containing the following for example:
Code:
This number 123456 is in line number one.
Line number two contains -333444 number.
The third line has a 111111 in between and also 435.
Line number four has :444444 in it.
Line five has a in front like a555555.

how can i filter out all the six-digit numbers in the text i.e the required output is:

Code:
123456
333444
111111
444444
555555

Thanks
# 2  
Old 07-13-2016
Hello james2009,

Could you please try following and let me know if this helps you. It will remove all 6 consecutive digits from each line.
Code:
awk  '{gsub(/[0-9]{6}/,X,$0);print}'   Input_file

EDIT: If you need to get as output from Input_file for all 6 consecutive digits then following may help you in same.
Code:
awk '{match($0,/[0-9]{6}/);print substr($0,RSTART,RLENGTH)}'   Input_file

Thanks,
R. Singh
Last edited by RavinderSingh13; 07-13-2016 at 11:56 AM.. Reason: Added possible solution
# 3  
Old 07-13-2016
Hi ravinder,

Thanks but it didn't work.
# 4  
Old 07-13-2016
I think the OP wants to see the numbers (works only with GNU tr command on Linux):
Code:
 awk  '{ for(i=1;i<=NF; i++)
          if( $(i) ~ /[0-9]{6}+/)
          {
            printf("%s\n", $(i) )
          }
       }'  filename | tr -dc '[[:digit:][:cntrl:]]'
123456
333444
111111
444444
555555

# 5  
Old 07-13-2016
Hi.

Also:
Code:
grep -E -o '[0-9]{6}' <file>

producing:
Code:
123456
333444
111111
444444
555555

on a system:
Code:
OS, ker|rel, machine: Linux, 3.16.0-4-amd64, x86_64
Distribution        : Debian 8.4 (jessie) 
bash GNU bash 4.3.30
grep (GNU grep) 2.20

Best wishes ... cheers, drl
This User Gave Thanks to drl For This Post:
james2009
# 6  
Old 07-13-2016
Thanks drl. It worked.
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