CODEWITHSUNDEEP
bashshellawksed
mezda
mezda

Reputation: 3647

passing command line argument to gawk script

I have a script chk.awk to which I want to pass some command line arguments. It has awk statements, sed command etc. Just for example I have taken a small program below to which I want to pass command line arguments.

#!/bin/bash
var1=$1
gawk '
BEGIN {
        printf "argc = %d\n argv0=%s\n argv1=%s\n var1=%s\n",ARGC,ARGV[0],ARGV[1],$var1
}'

But when I try :

$ sh chk.awk 10 20
argc = 1
argv0=gawk
argv1=
var1=

Above I tried to display the command line arguments by both ways i.e. argv & $1, but none of them work. Can anyone let me know where I am going wrong here? What is the correct way to do that?

Upvotes: 0

Views: 473

Answers (1)

Igor Chubin
Igor Chubin

Reputation: 64563

The problem is that you give arguments to the shell script, but not to the awk script. You must add "$@" to the call of gawk.

#!/bin/bash
var1=$1
gawk '
BEGIN {
    printf "argc = %d\n argv0=%s\n argv1=%s\n var1=%s\n",ARGC,ARGV[0],ARGV[1],$var1
}' "$@"

Otherwise you will your arguments in the shell-script and they will be not passed to gawk.

Update 1

If you have additional args (e.g. filenames that are to be processed), you must remove the first portition of args first (in the BEGIN section):

#!/bin/bash
var1=$1
gawk '
BEGIN {
    printf "argc = %d\n argv0=%s\n argv1=%s\n var1=%s\n",ARGC,ARGV[0],ARGV[1],$var1;
    delete ARGV[1]
}' "$@" filename

Upvotes: 1

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