Grep

donisback · #1 (**permalink**) 05-25-2009

Hi,
I have a file as:

Code:

IIN*00*00*ZZ*NDC *ZZ*11847*090520*0200*U*00401*000000001*0*P*>~GS*HP*NDC STD REMIT*1184722779*20090520*0200*63*X*004010X091A1~ST*835*10001~BPR*I*252480.19*C*CHK************20090515~TRN*1*0010855862*1952931460*12427008150981~DTM*405*20090514~N1*PR*CALIFORNIA*XV*H00~N3*ireland~

I want to extract 'CALIFORNIA' from the above one line file. The file will contain 'N1*PR*------' . I want to extract the string after N1*PR*.

Appreciate your help.

Regards,
Anil

pludi · #2 (**permalink**) 05-25-2009

If it's a fixed-field record with '*' as the separator, you could use

Code:

awk -F'*' '{print $49}' file

Otherwise, if you only want the part after 'N1*PR*', this Perl snippet might help:

Code:

perl -ne 'print $1,"\n" if /N1\*PR\*(.*?)\*/;' file

ghostdog74 · #3 (**permalink**) 05-25-2009

if you have Python

Code:

#!/usr/bin/env python
for line in open("file"):
    if "N1*PR*" in line:
        ind=line.index("N1*PR*")
        print line[ind+6:].split("*")[0]

output

Code:

# ./test.py
CALIFORNIA

vidyadhar85 · #4 (**permalink**) 05-25-2009

Hello,

Per our forum rules, all threads must have a descriptive subject text. For example, do not post questions with subjects like "Help Me!", "Urgent!!" or "Doubt". Post subjects like "Execution Problems with Cron" or "Help with Backup Shell Script".

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Thank you.

VIDYA

donisback · #5 (**permalink**) 05-26-2009

Thanks. Perl one-liner worked.

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