Passing arguments from one script to another

Question

Let's say I am in a Bash script and am sourcing a file. How do I pass variables to another script after I source the file like this.

sed -i 's/ = /=/' $file
source $file

Let's say file contains

variable1=10
variable2=apple

If I want to use these in another script, how do I pass these arguments to the other script, then run the script in my current Bash script.

Solution

You would pass them pretty much the same as you would pass arguments in any other way:

sed -i 's/ = /=/' "$file"
source "$file"

/path/to/another/script.sh "$variable1" "$variable2"

Obviously using the appropriate command line switches (or not if applicable).

If using the code as above, the value of $variable1 will be available in the other script as $1 (the 1st command line argument), while $variable2 will be available as $2.

To keep the original names in your new script you would need to reassign them using the positional parameters, ie:

variable1=$1
variable2=$2

However this may not be the most efficient way to do this, you might be better off with the suggestion below:

---

It sounds like you may actually want to source your file within the second script and not the first. In which case you may want to do the following:

script1.sh:

sed -i 's/ = /=/' "$file"
/path/to/another/script2.sh "$file"

script2.sh:

file=$1
source "$file"
printf '%s\n' "$variable1"
printf '%s\n' "$variable2"

Related recommended reading: 3.4.1 Positional Parameters

Note: assigning $1 to the file variable is not necessary, you could also simply source "$1" but I have written it this way in an attempt to show how positional parameters are handled

Comments

[αғsнιη]

please also read stackoverflow.com/q/5228345/4023950

[Admin]

I tried this (no necessary switches) and it's not working. The "another" script is running, because if I set the other script simply to "echo apple", it will display "apple", but if I set the other script to simply "echo $variable1", it displays a blank line. I know $variable1 has a value, because if I do "echo $variable1" before calling the other script, I get the proper value, so it seems that the variable is not getting passed.

[user394]

lol why is the last part hidden? Is it a spoiler? XD