Using an external variable inside a function

Question

I know title seems obscure but trust me I have researched this already but it fails for my code.

I have a code for licensing which pretty much looks like this

<?php if ( true ) : ?>
<?php print( 'Your License is Active' ); $type = 1; ?>

It pretty much sets the $type variable.

Now my problem:

function file_pull() {
if ( $type == 1 ) {
    $import_path = '/main/';

    $files_array = array(
        'start'               => array(),
        'home'                => array(
            'description' => __( 'Importing: Main Templates', 'kss' ),
            'files'       => array(
                'home.xml',
            ),
        ),

    );
}
if ( $type == 2 ) {
    $import_path = '/main/';

    $files_array = array(
        'start'               => array(),
        'home'                => array(
            'description' => __( 'Importing: Main Templates', 'kss' ),
            'files'       => array(
                'page1.xml',
            ),
        ),

    );
}

}

How can I use the $type variable inside this function so that it runs these file pull requests?

I have tried global to no avail. I have also tried setting it as a parameter.

function file_pull($type) {

Thanks.

1. Are you setting the $type and then calling the function in the same request? 2. Can you put var_dump($type); inside file_pull function (just above the first if) and post the result? — Goran
– Goran, Commented Mar 20, 2018 at 9:43
function file_pull($type) { is the way to go, but be aware that $type in this case does not automatically use your $type from outside the function. You'd need to call the function like file_pull(1) or file_pull(2) — user8034901
– user8034901, Commented Mar 20, 2018 at 9:48
Just to add on to @kerbholz's comment (which I also believe is the best solution to this question) $type is out of scope from your function. Have a read of the documentation on function scopes to get a better grasp on the issue. — IsThisJavascript
– IsThisJavascript, Commented Mar 20, 2018 at 9:56

user8034901 · Accepted Answer · 2018-03-20 10:38:28Z

1

Your function file_pull($type) { is a good start. Change your function to:

function file_pull($typeParam) {
    if ( $typeParam == 1 ) {
        $import_path = '/main/';

        $files_array = array(
            'start'               => array(),
            'home'                => array(
                'description' => __( 'Importing: Main Templates', 'kss' ),
                'files'       => array(
                    'home.xml',
                ),
            ),

        );
    }
    if ( $typeParam == 2 ) {
        $import_path = '/main/';

        $files_array = array(
            'start'               => array(),
            'home'                => array(
                'description' => __( 'Importing: Main Templates', 'kss' ),
                'files'       => array(
                    'page1.xml',
                ),
            ),

        );
    }
}

This way you can call it like:

$type=1;
file_pull($type);

I used $typeParam as function parameter to not get confused with your $type outside your function.

answered Mar 20, 2018 at 10:38

user8034901

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2 Comments

IsThisJavascript Over a year ago

Not sure who you're talking about when saying about confusion but PHP won't get confused if file_pull requires $type. The OP's code likely didn't work before because he probably didn't call file_pull($type);. Anyway, this wins my upvote as it's the solution I was hoping for.

user8034901 Over a year ago

I meant OP maybe getting confused about two $types there

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