Question

I’m struggling trying to read a php file inside a php and do some manipulation..after that have the content as a string, but when I try to output that with echo or print all the php tags are literally included on the file.

so here is my code:

function compilePage($page,$path){
   $contents = array();
   $menu = getMenuFor($page);
   $file = file_get_contents($path);
   array_push($contents,$menu);
   array_push($contents,$file);
   return implode("\n",$contents);
}

and this will return a string like

<div id="content>
   <h2>Here is my title</h2>
   <p><? echo "my body text"; ?></p>
</div>

but this will print exactly the content above not compiling the php on it.

So, how can I render this “compilePage” making sure it returns a compiled php result and not just a plain text?

Thanks in advance

Answer 1

function compilePage($page, $path) {
    $contents = getMenuFor($page);
    ob_start();
    include $path;
    $contents .= "\n".ob_get_clean();
    return $contents;
}

To evaluate PHP code in a string you use the eval function, but this comes highly unadvised. If you have a file containing PHP code, you can evaluate it with include, include_once, require, or require_once depending on your need. To capture the output of an included file – or required, or whichever method – you need to enable output buffering.