custom Switch button in footer

[craigfeldspar]

I've made a test page with the same footer operation, let me know if this is better for you: https://benoitlefeuvre.com/test/

thanks!

[craigfeldspar]

hi
up 🧌

[arminunruh]

hey oh yea i need to check this out sry

[craigfeldspar]

in this case, the switch button should be on Artworks, as the page category is data-catid=“[3]” and should display the correct content as if Artworks had been clicked.

[arminunruh]

im checking it out now

[arminunruh]

ok it seems like i had a really hard time to understand what the problem was :O
i misunderstood you

basically you just need to trigger a click on one of the buttons

u got two buttons:

.artworks-button

and

.overview-button

and as far as i understand this would be the way to do it:

when you go to a page where you want overview-button to be auto-activated:

window.laytheme.on('newpageshown', function(){
if(jQuery("body").hasClass("slug-home"){
jQuery(".overview-button").click();
}
})

basically when you visit the homepage, this will click the overview button using javascript:

but maybe u need to remove this custom js u got here:

this could be in conflict with that click js
i think u just add "selected" class

maybe for styling, instead use the class "expand-link-clicked" cause thats the "active" class

so basically with newpageshown and checking the body class you can find out which page u just landed on.
and then either call

jQuery(".overview-button").click();

or

jQuery(".artworks-button").click();

if my previous code was checking the right pages (no idea), it could for example look like

window.laytheme.on('newpageshown', function(){
	if(jQuery('body').attr('data-catid') == '[3]'){
		jQuery(".artworks-button").click();
	} else if(jQuery('body').attr('data-catid') == '[1]'){
		jQuery(".overview-button").click();
	}
})

but basically this was no lay theme bug, you just needed help with your custom code

[craigfeldspar]

Thanks for your reply. You're right, it's probably not a Lay bug but rather that I need help with my code. I tested it, but unfortunately it doesn't work.

[arminunruh]

:/
can you tell me:
which page should have what button activated??

and can u gimme your /wp-admin/ login

can you send your website address, /wp-admin/ username and password and a link to this topic to info@laytheme.com?

[arminunruh]

<script>
    window.laytheme.on('newpageshown', function(){
        setTimeout(function(){
        	if(jQuery('body').attr('data-catid') == '[3]'){
        	    if( window.utility.isDesktopOrTabletSize ) {
            		jQuery("#footer .artworks-button").click().addClass("selected");
            		jQuery("#footer .overview-button").removeClass("selected");
        	    } else {
        	        jQuery("#footer-custom-phone-grid .artworks-button").click().addClass("selected");
        	        jQuery("#footer-custom-phone-grid .overview-button").removeClass("selected");
        	    }
        	} else if(jQuery('body').attr('data-catid') == '[1]'){
        	    if( window.utility.isDesktopOrTabletSize ) {
        		    jQuery("#footer .overview-button").click().addClass("selected");
        		    jQuery("#footer .artworks-button").removeClass("selected");
        	    } else {
        	        jQuery("#footer-custom-phone-grid .overview-button").click().addClass("selected");
        	        jQuery("#footer-custom-phone-grid .artworks-button").removeClass("selected");
        	    }
        	}
        }, 1000)
    })
</script>

seems like this code is what was needed

[craigfeldspar]

Thank you very much for your help and your time. That's exactly what I was looking for 🙏