[Resolved] Conditional Test For Disabled Elements

[Antar]

Hello, All –

Is there a way to test if certain elements have been disabled? I want to create hooks that test if (for example) the featured image is disabled, or the title, etc.

How should I do this?

Thanks,

Antar

[Tom]

You could do this:

$disable_top_bar = get_post_meta( get_the_ID(), '_generate-disable-top-bar', true );
$disable_header = get_post_meta( get_the_ID(), '_generate-disable-header', true );
$disable_nav = get_post_meta( get_the_ID(), '_generate-disable-nav', true );
$disable_secondary_nav = get_post_meta( get_the_ID(), '_generate-disable-secondary-nav', true );
$disable_post_image = get_post_meta( get_the_ID(), '_generate-disable-post-image', true );
$disable_headline = get_post_meta( get_the_ID(), '_generate-disable-headline', true );
$disable_footer = get_post_meta( get_the_ID(), '_generate-disable-footer', true );

Then check like this:

if ( $disable_top_bar ) {
    // Top bar is disabled
}

[Antar]

Perfect. Just what I was hoping for.

Thanks, Tom.

Antar

[Tom]

You’re welcome 🙂

[Elvis]

Hello,
I want to have my site footer be a <footer> element and located direcly inside body.
So, I clicked on my default element > layout to dissable site-footer.
Then created my footer as an element > block > hook with before_footer location.
And it works.
In order to clean after my self, I wish to remove the now empty <div class="site-footer"> </div> markup. But I wish to run that only when the dissable footer option is actually ran.

Tried this:

function remove_site_footer_wrapper() {
    $disable_footer = get_post_meta( get_the_ID(), '_generate-disable-footer', true );
    if ( $disable_footer ) {
	?>
		<script type="text/javascript">
			const siteFooter = document.querySelector( ".site-footer" );	
			siteFooter.parentNode.removeChild( siteFooter );
		</script>
		<style>
			/** If no JS */
			.apuri-body > .site-footer {display: none; visibility: hidden;}
		</style>
	<?php
    }
}
add_action('wp_footer', 'remove_site_footer_wrapper');

If I remove $disable_footer condition, it works, with it id does not! What am I missing?
Thanks.

[David]

Hi there,

how is the Element being disabled ? As I believe the above only applies if you use the Disable elements meta box in the post editor.

[Elvis]

It’s dissabled via option in the layout element that is acive globally, with some exceptions. It would make no sense to dissable a global site part such as site-footer via individual page, if I plan not to use it on most pages. But yeah get_the_ID() does imply that what you are saying is true.

So there are no alternative ways to go about this?
Thanks

[David]

See Toms reply here:

https://generatepress.com/forums/topic/how-to-hook-into-if-a-post-of-page-is-using-an-element-to-conditionally-show-acf/#post-1747328

[Elvis]

O, nice. 🙂 Thanks. So we can also load assets for element only when element is present. Yummy. 🙂
Can this be done with local templates too?!

[Elvis]

Ok, I can do it, but kinda crudely.

function remove_site_footer_wrapper() {
	$disable_footer = get_post_meta( get_the_ID(), '_generate-disable-footer', true );
	global $generate_elements;
	if ( $generate_elements[144] ) {
	?>
		<script type="text/javascript" id="apuri-default-footer-remove">
			const siteFooter = document.querySelector( ".site-footer" );	
			siteFooter.parentNode.removeChild( siteFooter );
		</script>
		<style>
			/** If no JS */
			.apuri-body > .site-footer {display: none; visibility: hidden;}
		</style>
	<?php
	}
}
add_action('wp_footer', 'remove_site_footer_wrapper');

But what I would like to check is weather the block in question has the dissable footer true. I tried this, but no luck. 🙂

function remove_site_footer_wrapper() {
	global $generate_elements;
	foreach ( $generate_elements as $element ) {
	$disable_footer = get_post_meta( $element->ID, '_generate-disable-footer', true );
		if ( $disable_footer ) {
		?>
			<script type="text/javascript" id="apuri-default-footer-remove">
				const siteFooter = document.querySelector( ".site-footer" );	
				siteFooter.parentNode.removeChild( siteFooter );
			</script>
			<style>
				/** If no JS */
				.apuri-body > .site-footer {display: none; visibility: hidden;}
			</style>
		<?php
		}
	}
}
add_action('wp_footer', 'remove_site_footer_wrapper');

the print_r($generate_elements); shows id and element type. So cannot get it from there.
Thanks a lot.

[Tom]

Try this as the foreach:

foreach ( $generate_elements as $key => $data )

And this to get the ID: $data['id']

[Elvis]

Thanks Tom,
sure it’s an array [facepalm].

Tried but does not work. I am able to get the ID both with $generate_elements as $data as well as $generate_elements as $key => $data when I use $data['id'] to output.

function prefix_get_your_elements() {
	global $generate_elements; 
	foreach ( $generate_elements as $data ) {
	?>
	<pre style="min-height: 800px; padding-top: 7em">
		<?php print_r($data['id']); ?>
	</pre>
<?php 
	}
}

This outputs the id’s of elements used on the current page. But, the link between the element (identified by the id) and the option to dissabled footer is where it fails.

This is my function

function remove_site_footer_wrapper() {
	global $generate_elements;
	foreach ( $generate_elements as $key => $data ) {
	$disable_footer = get_post_meta( $data['id'], '_generate-disable-footer', true );
		if ( $disable_footer ) {
		?>
			<script type="text/javascript" id="apuri-default-footer-remove">
				const siteFooter = document.querySelector( ".site-footer" );	
				siteFooter.parentNode.removeChild( siteFooter );
			</script>
			<style>
				/** If no JS */
				.apuri-body > .site-footer {display: none; visibility: hidden;}
			</style>
		<?php
		}
	}
}
add_action('wp_footer', 'remove_site_footer_wrapper');

And it does not work.
Thanks

[Fernando]

Hi Elvis,

You can try modifying your code into something like this:

<?php 
global $generate_elements;
foreach ( $generate_elements as $generate_element ) {
		if ( $generate_element['type'] === "layout" && $generate_element['id'] === 391895 ) {
		?>
			<script type="text/javascript" id="apuri-default-footer-remove">
				const siteFooter = document.querySelector( ".site-footer" );	
				siteFooter.parentNode.removeChild( siteFooter );
			</script>
			<style>
				/** If no JS */
				.apuri-body > .site-footer {display: none; visibility: hidden;}
			</style>
		<?php
		}
	}
?>

This would check if there is a Layout Element with a specific ID active on a page.

Kindly replace 391895 with the ID of your Layout Element disabling the Site Footer.

Try to var_dump($generate_elements) to get the specific id of the Layout Element.

Tested this on my end and it seems to be working as expected.

Kindly let us know how it goes. 🙂

[Elvis]

Hello Fernando,

thanks for this, I have something like this and it works, but it’s hardcoding a particular element.
What I wanted is to really check for the dissable option, because then I would have a portable code block to use accross my sites, that would work whenever a layout block dissables a footer, header … or anything.

Thanks again.

[David]

Hi there,

maybe a curve ball – what about using has_action()

https://developer.wordpress.org/reference/functions/has_action/

to determine whether the hook has the relevant callback.

has_action('generate_footer', 'generate_construct_footer')

[Author]

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