concerning a page you can use
$exclude=stristr(‘https://’ . $_SERVER[‘HTTP_HOST’]. $_SERVER[‘REQUEST_URI’],’/some-page-name/’);
if(!$exclude {…}
The advance of this solution is that it concerns also sub-pages like /some-page-name/sub-page-name/
apparenty WordPress has also some functions for getting the page id or post id. Look at function reference
https://codex.wordpress.org/Function_Reference
$exclude1 = some_function_name(‘here page/post id’);
$exclude2 = some_function_name(‘here page/post id’);
if (!($exclude1 || $exclude2)) {…}
Thread Starter
Andrzej
(@ukandrzej)
Thank you very much.
I understand where I need to substitute my page URL but don’t understand what I need to put in the curly brackets: {…}.
Am I able to add this code above or below into the same code snippet that I want to work on the rest of the site?
Inside curly brackets just the code, which you want to be executed with other pages.
Hello @ukandrzej,
I would recommend adding this to the top of snippets that you want to exclude from a specific page:
if ( is_page( 'page-slug' ) ) return;
You can also pass in page IDs or titles, or multiple pages in an array. See more here: https://developer.wordpress.org/reference/functions/is_page/
Thread Starter
Andrzej
(@ukandrzej)
Hi Shea,
I’m adding your code like this:
if ( is_page( '/categories/hotels/' ) ) return;
function c3_display_dbl_spaces($content){
return preg_replace("/(.|\?|!)( {2})/", "\\1 ", $content);
}
add_filter( 'the_content', 'c3_display_dbl_spaces', 20 );
The php snippet is to represent a non breaking space which I add with the space bar after a sentence full stop. It works fine but not on the page path mydomain.com/categories/hotels/ (page ID is 17493), where it breaks my external affiliate JavaScript code which is supposed to pull in up to date information from the business.
Hi Andrej,
You need to pass just the slug to is_page() – not the full path.
It might be easier to use the ID in this situation, like this:
add_filter( 'the_content', function ( $content ) {
if ( is_page( 17493 ) ) {
return $content;
}
return preg_replace( '/(.|\?|!)( {2})/', '\\1 ', $content );
}, 20 );
Thread Starter
Andrzej
(@ukandrzej)
Hi Shea,
I very much appreciate the help from you both. I used your code above and it works perfectly.
Best regards,
Andrzej
