variable passed to awk

whatisthis · #1 (**permalink**) 09-14-2004

Does anybody know how to print a variable passed to awk command?

awk -F"|" 'BEGIN {print $job,"\n","Question \n"} {print $1,$2$4," ",$3}' "job=$job1" file1

I am trying to pass job the variable job1.

the output is blank.

??

google · #2 (**permalink**) 09-14-2004

Use the -v switch to pass in a variable to Awk. Go to ]GNU and search for gAwk documentation for more details.

Code:

awk -v job=$job BEGIN '{FS= "|"; print "$job\n","Question \n"} {print $1,$2,$4," ",$3}'  file1

whatisthis · #3 (**permalink**) 09-15-2004

It's still not working.
variable $job is not empty string. but when it printed, it always prints JOB $job

echo job is $job
nawk -F"|" -v JOB=$job 'BEGIN {print "JOB ","$job\n"}
{print $1,$2," ",$4,"\n"}' list

google · #4 (**permalink**) 09-15-2004

Thats because you are using "print" and placing job in quotes. Try using printf instead and using the %s notation.

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