PHP“如果'任何'这些选项设置，显示它

Question

So I'm using WordPress for this project, and I'm not quite sure how to go about this neat & clean.

I've got 8 options total. If ANY of them are set, I need something to happen. I also need to execute it for each one set.

Let's use two, for now

if( (genesis_get_option('setting_one', 'settings') || genesis_get_option('setting_two', 'settings')) !== '')
    { // print() each one that is set }

You can see how doing it that way for 8 options would get very ugly very quickly. All of my settings all have an identical prefix: (setting_a, setting_b, setting_one,setting_two)

I was thinking something like

$settings = array_fill(1, 8, 'setting_');
$labels = array(1 => 'a', 'b', 'c', 'd', 'one', 'two', 'three', 'four');

.... but then I don't know where to go from there.

There's GOT to be some sort of clean way to do this?

Answer 1

A somewhat neater way would be:

$settings = array("setting_one","setting_two","setting_a");
$has_setting = false;
foreach($settings as $setting){
    if(genesis_get_option($setting, 'settings') !== ''){
        $has_setting = true;
        break;
    }
}
if($has_setting){
    //Do stuff
}

Even better would be putting the check into a method to keep the code neater.

Edit: Just realised you need an action for each setting. In that case you can do it inside the loop.