Conditionally show grid?

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Is there any way of only showing a grid if it's filled with content? I'm settings up the pages so that you either can choose to simply put the content in the content editor OR choose to create a grid with LayGridder – or both. I'm showing the grid in the page.php template with <?php the_laygrid($id, $type); ?>.

But at the moment, an empty div for LayGridder is displayed regardless of if the grid has any content or not, which affects the layout a bit since the empty div is taking up a bit of space.

So I'm wondering if there's any technique that you can use to see if the grid has any content and then conditionally output the markup for it depending on that boolean?

You could go the other way around. Check if there is content, if yes show the content, otherwise show grid.

/* if content is not empty, show it. else show laygrid */
if(get_the_content() != '' ) {
    the_content();
} else {
    the_laygrid();
}

I am just a user. So maybe the dev-team has a better idea for a conditional tag. The other way around would be better for fallback. I wrap my function with a check if laygridder is a active plugin otherwise show the_content (code is here)

If you have Advanced Custom Fields or similar stuff installed you could also do a conditional field in the backend post editor what editor should be escaped in the frontend.

Dear @tobiasger
there is not a default way. But you could check with PHP or jQuery if there is content and if not – remove it.

Best!

Marius

You could set a boolean to show/hide Laygridder in PHP with the following:

$gridder_json = get_post_meta( $id, '_gridder_json', true );
$obj = json_decode($gridder_json);
$show_laygridder = (!empty($obj->cont)) ? true : false;

If the variable $show_laygridder is true, show Laygridder:

if($show_laygridder):
  the_laygrid($id);
endif;

It's just a simple solution, it won't catch empty text elements etc.

Dear @hb
yes! Nice!

Best!

Marius