How to pass file path to a function as an argument in bash script?

Question

I want to take file from different directory and pass it to a function as an argument. I have been at this for hours but I can not get it to work. So far I have tried this

#!/bin/bash

myfile=/home/$USER/Desktop/Programs/Files/asd.bam

 function mate()
{ 
    samtools index $arg

    echo "$arg"
 }

mate $myfile

I check argument with echo command but It shows empty.

Thanks in advance.

Have you really been searching on this topic? Even a simple [google search](https://www.google.pl/search?q=bash+function+arguments) gives a lot of info? — KamilCuk, Apr 09 '19 at 11:41
Yes I have and I wrote, I have been trying to get it work for hours. Maybe I am dim-witted or something. — Saruman, Apr 09 '19 at 11:44
Hm, ok. Just replace `$arg` with `$1`. Ex `echo "$1"`. And it will work. — KamilCuk, Apr 09 '19 at 11:50

JRFerguson · Accepted Answer · 2019-04-09T16:05:00.230

1

You haven't defined arg and hence its is empty (undefined). $1 within the function will be the first argument passed.

function mate()
{ 
    arg="$1"
    samtools index "$arg"
    echo "$arg"
}

edited Apr 09 '19 at 16:05

answered Apr 09 '19 at 11:49

JRFerguson

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How to pass file path to a function as an argument in bash script?

1 Answers1