[PHP] How to assign eval() to variable?

[Afan Pasalic]

hi,
I have on one website boxes with information, pulled from mysql. the 
content can be string, php code, url of other website or url to specific 
file etc.

currently, I have something like this:

// connect to db
// mysql_query() to get box content and content_type

switch($content_type)
{
  case 'string':
     echo $content;
     break;

  case 'php_code':
     eval($content);
     break;

  case 'website':
     echo '<iframe>'.$content.'</iframe>;
     break;

  case 'file'
     require_once($file);
     echo $file_content;

  // etc.
}

but, now I have to change the code to assign content to variable and the 
variable will be printed later. I tried something like this:


switch($content_type)
{
  case 'string':
     $record = $content;
     break;

  case 'php_code':
     $record = eval($content);
     break;

  case 'website':
     $record = '<iframe>'.$content.'</iframe>;
     break;

  case 'file'
     require_once($file);
     $record = $file_content;

  // etc.
}

and it works - except eval() part. cant do $record = eval($content); 
?!?!?!?

thanks


afan







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