Pass parameter to an awk script file

Question

If I want to pass a parameter to an awk script file, how can I do that ?

#!/usr/bin/awk -f

{print $1}

Here I want to print the first argument

Answer 1

your hash bang defines the script is not shell script, it is an awk script. you cannot do it in bash way within your script.

also, what you did : echo blah|awk ... is not passing paramenter, it pipes the output of echo command to another command.

you could try these way below:

 echo "hello"|./foo.awk file -

or

var="hello"
awk -v a="$var" -f foo.awk file

with this, you have var a in your foo.awk, you could use it.

if you want to do something like shell script accept $1 $2 vars, you can write a small shellscript to wrap your awk stuff.

EDIT

No I didn't misunderstand you.

let's take the example:

let's say, your x.awk has:

{print $1}

if you do :

echo "foo" | x.awk file

it is same as:

echo "foo"| awk '{print $1}' file

here the input for awk is only file, your echo foo doesn't make sense. if you do:

  echo "foo"|awk '{print $1}' file -
or
    echo "foo"|awk '{print $1}' - file

awk takes two input (arguments for awk) one is stdin one is the file, in your awk script you could:

echo "foo"|awk 'NR==FNR{print $1;next}{print $1}' - file

this will print first foo from your echo, then the column1 from file of course this example does nothing actual work, just print them all.

you can of course have more than two inputs, and don't check the NR and FNR, you could use the

ARGC   The number of elements in the ARGV array.

ARGV   An array of command line arguments, excluding options and the program argument, numbered from zero to ARGC-1

for example :

echo "foo"|./x.awk file1 - file2

then your "foo" is the 2nd arg, you can get it in your x.awk by ARGV[2]

echo "foo" |x.awk file1 file2 file2 -

now it is ARGV[4] case.

I mean, your echo "foo"|.. would be stdin for awk, it could by 1st or nth "argument"/input for awk. depends on where you put the -(stdin). You have to handle it in your awk script.