Shell Programming and Scripting

Try:
grep '.\{44\}00[23].*' infile > outfile

Thanks perderabo, but what if the strings were completely different ie, didnt have a common 00 at the front. I have another status code of 55 (with a space after to make up the 3rd character ie

xxxxxx,002,xxx
xxxxxx,003,xxx
xxxxxx,55 ,xxx

I cant really use the '00' as a prefix and I presume I cant put [55002003] into the command ??

I have been advised by somebody else to use | inside some () to get multiple values into the condition ie

grep '.\{44\}\(002|55\).*' infile > outfile

although this doesnt seem to work, does anybody know if im on the right track ??

ive tried it with egrep instead of grep and also without escaping the ()'s all to no avail

any help would be greatly appreciated

I'm starting to think that awk would solve these problems easier.

You've got 7 comma seperated fields in the input file.

The "status" is in the 6th field.

Therefore

Code:

awk 'BEGIN { FS="," } { if ( $6 ~ /55|002|003/ ) { print $0 } }' infile > outfile

Would place any record with status 55, 002 or 003 into "outfile".

Modify the ( $6 ~ /55|002|003/ ) part and insert the statuses (or is that statii?!) of the records you want.

Cheers
ZB

works perfectly thankyou, out of interest how would i do a "not" version of this script so that everything other than 002,003,55 get put in a file

cheers

Change the if statement to
... if ( $6 !~ /55|002|003/ ) ...

Note: ~ means "matches"
!~ means "doesn't match"

Cheers
ZB

Hi,
I am using the following AWK command to strip some lines from a file based on a pattern. But this is creating an additional blank line at the position where the pattern matched. Please advise how to avoid the blank line. Thanks.

awk 'BEGIN { FS="," } { if ( $6 ~ /55|002|003/ ) { print $0 } }' infile > outfile

Regards,
Tipsy.