The issue in your script is that myVariable is defined inside the function my_test(), so its scope is limited to that function. To make the variable accessible outside the function, you need to declare it as global or ensure it is set in the global scope.

Solution 1: Declare myVariable as global inside the function

bash
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echo "myVariable: $myVariable|"

my_test() {
    local myVariable=true  # This makes it local to the function
    # To make it global, remove 'local' or explicitly declare it as global
    myVariable=true
}

my_test
echo "myVariable: $myVariable|"

However, by default, variables in Bash are global unless declared as local. So if you remove local, it will work as expected.

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Solution 2: Set myVariable globally before the function

bash
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myVariable=false  # Set globally

echo "myVariable: $myVariable|"

my_test() {
    myVariable=true  # This will modify the global variable
}

my_test
echo "myVariable: $myVariable|"

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Solution 3: Use export to make it available in subshells (if needed)

If you want the variable to be available in subshells, you can use export:

bash
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export myVariable=false

echo "myVariable: $myVariable|"

my_test() {
    myVariable=true
}

my_test
echo "myVariable: $myVariable|"

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Final Working Example

Here’s the corrected version of your script:

bash
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echo "myVariable: $myVariable|"

my_test() {
    myVariable=true
}

my_test
echo "myVariable: $myVariable|"

Output:

text
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myVariable: |
myVariable: true|

The key is to call the function (my_test) before the second echo. Otherwise, myVariable remains unset.