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Home/ Questions/Q 6742677
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T11:52:18+00:00 2026-05-26T11:52:18+00:00

I have the following code which works fine inline with my code: if ($progressData[1]

  • 0

I have the following code which works fine inline with my code:

    if ($progressData[1] == "yes") {

        echo "Complete";
    }
    else

        echo "Not Yet Complete";

However, I would like to call it from a function:

function progressOutput () {

    if ($progressData[1] == "yes") {

        echo "Complete";
    }
    else

        echo "Not Yet Complete";

}   

When I call progressOutput(), I get “Not Yet Complete”, even though $progressData[1] is equal to “yes”.

Here is how I am calling the function:

Mission Status: <?php progressOutput(); ?>

What do I need to do to get progressOutput() to return “Complete” when $progressData[1] is in fact equal to “yes”?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T11:52:19+00:00Added an answer on May 26, 2026 at 11:52 am

    You need to pass the variable from the calling scope into the function in order for the function to have access to it.

    Your function should accept an argument:

    function progressOutput ($progressData) {
        if ($progressData[1] == "yes")
            echo "Complete";
        else
            echo "Not Yet Complete";
    }   
    

    And when you call it, you should provide an argument:

    Mission Status: <?php progressOutput($progressData); ?>
    

    It’s typically a bad idea for your functions to output data directly; you could clean it up by returning a value instead of echoing it:

    function progressOutput ($progressData) {
        if ($progressData[1] == "yes")
            return "Complete";
        return "Not Yet Complete";
    }  
    

    And outputting the value returned by the function:

    Mission Status: <?= progressOutput($progressData); ?>
    
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