I am not able to concatenate variable in php

Question

I am making some changes to an existing Wordpress theme.

public function checkTv( $post ) {
         global $title;
        if ( ! empty( $post['season'] ) ) { 
            $videourl ='shows'.$title. $post['season'].'-'. $post['episodio'];
        }

        return $videourl;
    }

Here the $videourl contains the desired URL format. Everything is working but the $title value is not being concatenated in the URL. It is being skipped automatically. In title i have the slug.

This how the call is being made

$postmeta = doo_postmeta_episodes($post_id);
$videourl = $this->checkTv( $postmeta );

$title has been declared as global and the value of the title is being taken from a function.

Possible duplicate of [Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors?](https://stackoverflow.com/questions/16959576/reference-what-is-variable-scope-which-variables-are-accessible-from-where-and) — Artem Ilchenko, Mar 14 '19 at 07:58
_“$title has been declared as global”_ - no it has not, that would require the `global` keyword _inside_ your function. — 04FS, Mar 14 '19 at 08:01

score 2 · Answer 1 · answered Mar 14 '19 at 08:05

2

PHP does not skip the $title variable: It is undefined in your function and therefore empty. To use a variable which is defined outside the function you need to put the global directive inside of your function:

public function checkTv($post) {
    global $title;
    //...
}

This informs the function that the $title you're about to use is the same as the one declared outside the function.

answered Mar 14 '19 at 08:05

wortwart

3,139
27
31

sir its still not concatenating $title to the $videourl. Can you help me in this ? – Robot Mar 14 '19 at 08:53

Emre Erdoğan · Answer 2 · 2019-03-14T09:20:29.207

0

Try this:

$videourl = 'shows'
    . get_the_title($post) 
    . '-' 
    . $post['season'] 
    . '-' 
    . $post['episodio'];

UPDATE: Based on the comment that you stated you need slug instead of title, try this:

$slug = get_post_field('post_name', $post);
$videourl = 'shows' 
    . $slug
    . '-' 
    . $post['season'] 
    . '-' 
    . $post['episodio'];

edited Mar 14 '19 at 09:20

answered Mar 14 '19 at 08:01

Emre Erdoğan

119
4

sir this works but its giving the title as hello world. I need slug. Which is in url. – Robot Mar 14 '19 at 08:56
hey can you check my code with anydesk ? it would be a great help – Robot Mar 14 '19 at 09:30

score 0 · Answer 3 · answered Mar 14 '19 at 09:44

0

Thanks for your help. I solved it. The Problem was with the scope but declaring it global was not solving it. I passed a another parameter with $post in the function that I got from $post_id and it solved the problem. thanks for your efforts

answered Mar 14 '19 at 09:44

Robot

1
3

I am not able to concatenate variable in php

3 Answers3