Re: How to assign eval() to variable?

short hack works like a charm! :-) thanks! afan oorza2k5@gmail.com wrote:

short hack, assuming your eval echo's out to the browser (which I can't see how else you'd expect something to 'return' from an eval'd statement ob_start(); eval($foo); $result = ob_get_clean(); On May 23, 2009 1:46am, Afan Pasalic <afan@afan.net> wrote: hi, I have on one website boxes with information, pulled from mysql. the content can be string, php code, url of other website or url to specific file etc. currently, I have something like this: // connect to db // mysql_query() to get box content and content_type switch($content_type) { case 'string':

    echo $content;

    break;

case 'php_code':

    eval($content);

    break;

case 'website':

    echo ''.$content.';

    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':

    $record = ''.$content.';

    break;

case 'file'

    require_once($file);

    $record = $file_content;

// etc. } and it works - except eval() part. cant do $record = eval($content); ?!?!?!? thanks afan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php