Question

In my code, I use a public load_snippet function of a class when I need to include HTML or PHP snippets. (I do this instead of a direct include_once because the directory structure varies depending on certain variables).

I had some issues with variable scopes, so I’ve narrowed down the problem to this: let’s say I define a variable within my page:

$variable = 'Hello World!";

Then, I need load in a snippet:

$APP->load_snippet("slider");

The snippet renders perfectly, except that PHP gives an undefined variable error if I try to reference $variable in the slider code. If I directly include the php file, it works as expected, so I don’t understand why I’m having this problem, since this is the load_snippet function:

public function load_snippet($snippet){  
if(file_exists("app/".$this->APP_TYPE."/snippets/".$snippet.".php")){  
        include "app/".$this->APP_TYPE."/snippets/".$snippet.".php";  
    }  
    else{  
        include 'common/txt/404.txt';  
    }  
}

Any help you can give me is much appreciated.

Answer 1

The file is being included within the context of the load_snippet() function, and therefore has only those variables which exist within that function. One way to modify this is to make your function accept two variables: the filename and an array of values.

public function load_snippet($snippet, $content) {  
    if (is_array($content)) extract($content);
    if (file_exists("app/".$this->APP_TYPE."/snippets/".$snippet.".php")) {  
        include "app/".$this->APP_TYPE."/snippets/".$snippet.".php";  
    } else {  
        include 'common/txt/404.txt';  
    }  
}

Then

$arr = array('variable' => 'Hello world!');
load_snippet('slider', $arr);