Pass command line arguments to awk

Question

I am trying to make a simple bash script that uses command line arguments as parameters to awk.

The file format is like this:

id|Name|birthday|joinDate

And the command line input:

./my_script -f filename.dat -id 1234

When I use the following line of code:

awk -v arg ="$4", -F"|" '{if(arg == $1)print $1,$2}' "$2"

I get an error that says:

'awk: 'arg' argument to '-v' not in 'var=value' form'.

What's wrong with this? All I want is to print specific columns of the given ID.

karakfa · Accepted Answer · 2016-06-29 15:42:18Z

9

Your answer is in the error message. Put the argument in var=value form, without any spaces around = sign.

answered Mar 13, 2016 at 16:54

karakfa

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