PHP：如何在另一个函数中检查函数的返回值

Question

I have a function which returns a value that I'd like to check via another one. I can not echo it in the first function as far as I know, as it's used in a wordpress filter for body_class and outputted there.

So, how do I check what the returned value is in another function?

An example of the first function that returns $class, which is what I want to check:

function layout_class( $class ) {
    // There is lots of more functionality here for the $layout variable
    $layout = 'col-3cm';

    $class[] = $layout;
    return $class;
}
add_filter( 'body_class', 'layout_class' );

Now, this class decides whether or not to load a secondary sidebar template. So I would like to do:

function sidebar_dual() {
    $layout = layout_class();   
    if (
        ( $layout == 'col-3cm' ) ||
        ( $layout == 'col-3cl' ) ||
        ( $layout == 'col-3cr' )
    )
    { get_template_part('sidebar-2'); }
}

Since I can't echo out the first function, and don't want to write another function as it is quite big - how do I approach it? Is there an easy way to check a return value, similar to echo?

Answer 1

It looks like you need to refactor your code.

You just need to move the code that determines the layout into a separate function that you can call both inside the filter and inside the sidebar_dual() function:

function get_layout_mode(){

   // There is lots of more functionality here for the $layout variable
   $layout = 'col-3cm';

   return $layout;
}

The filter function becomes:

function layout_class( $class ) {
    $class[] = get_layout_mode();
    return $class;
}
add_filter( 'body_class', 'layout_class' );

And in your sidebar_dual() function call get_layout_mode() instead of layout_class()

---

It is possible to work with your current code too, by splitting the string returned by WP's get_body_class() function into an array then checking if any of those 3 classes are present inside the array. I would go with the first option though, because it's clearer.