BASH Run multiple scripts from a another script

Question

I have a doubt about running multiple scripts from a third one:

first.sh

#!/bin/bash
echo "script 1"
#... and also download a csv file from gdrive

second.sh

#!/bin/bash
echo "script 2"

third.awk

#!/usr/bin/awk -f 

BEGIN {
    print "script3"
}

I would like a 4th script that run them in order, I've tried the following but only runs the first script.

#!/bin/bash

array=( first.sh second.sh )
for i in "${array[@]}"
do
   chmod +x $i
   echo $i
   . $i 
done

But only runs the first script and nothing else.

Thank you very much for the support! Santiago

Loop through the index or the array, something like: for i in "${!array[@]}"; do echo "${array[i]}}"; done — Jetchisel
– Jetchisel, Commented Jan 12, 2023 at 15:11

choroba · Accepted Answer · 2023-01-12 17:09:09Z

1

You can't source an awk script into a shell script. Run the script instead of sourcing it.

. (aka source) executes commands from the file in the current shell, it disregards the shebang line.

What you need instead is ./, i.e. path to the script, unless . is part of your $PATH (which is usually not recommended ).

#!/bin/bash

array=( first.sh second.sh )
for i in "${array[@]}"
do
   chmod +x "$i"
   echo "$i"
   ./"$i"  # <---
done

Why is the second script not running? I guess the first script contains an exit, which when sourced exits the shell, i.e. it doesn't continue running the outer wrapper.

edited Jan 12, 2023 at 17:09

answered Jan 12, 2023 at 15:26

choroba

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3 Comments

Ivan Over a year ago

This "$i" is not working at all. The dot is needed but with a "/", like that ./"$i", see?)

choroba Over a year ago

@Ivan: It depends on the setting of $PATH, but generally, you're right. Fixed.

choroba Over a year ago

But ./ is a path, while . with a space is a builtin command.

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