topic Re: awk in Operating System - HP-UX

awk

Gary Price — Fri, 11 Aug 2000 17:16:32 GMT

I am trying to use awk to print fields of a log file {print $5 $6}. The area where I need assistance is printing the remaining record/entry. The log file is dynamic and I need to display fields 5,6 and everything after six.

Thanks,
Gary

Re: awk

Scott D. Allen — Fri, 11 Aug 2000 17:26:47 GMT

use NF to give yourself the number of fields in each record and loop it or something like that.

Re: awk

Kofi ARTHIABAH — Fri, 11 Aug 2000 17:27:04 GMT

I forget the awk construct for this but this should do the same:

# cut -d delimiter -f 5,6- filename

where delimiter is the field delimiter and filename is the file you want to process.

Good luck

Re: awk

James R. Ferguson — Fri, 11 Aug 2000 17:44:11 GMT

Gary:

Use:

awk '{for (i=5; i <= NF; i++1) print $i}'

where i is initialized to the first field you want!

...JRF...

Re: awk

Anthony deRito — Fri, 11 Aug 2000 17:46:32 GMT

Gary, I am guessing that you want to set some variables here.

VAR1=[field 5],
VAR2=[field 6],
VAR3=[everything_after_six]

You can use field specifiers to process each record like this:

cut -d "delimiter" -f 5,6

Use this to set VAR1 and VAR2

VAR3 can be set also by using cut as well but use the "-" range specifier like this:

cat -d "delimiter" 7-

This will cut positions 7 through the end of the record and thus can use it to set VAR3.

Tony

Re: awk

Anthony deRito — Fri, 11 Aug 2000 17:48:34 GMT

Gary, sorry for the typo, the last command should be

cut -d "delimiter" 7-

not

cat -d "delimiter" 7-

Tony

Re: awk

Kofi ARTHIABAH — Fri, 11 Aug 2000 18:20:14 GMT

Gary:

Just out of curiosity, could you run "time" against the two versions - I am curious to find out which does the job quicker - cut or awk.

Thanks