Hello landers

Am trying to include a php page when function is been call so i use this code



document.getElementById("content".innerHTML = "<?php include('file.php'); ?>";

It didn't work for me.

pls any help will be grateful

it's not ideal to do stuffs like that.when u include files from my php knowledge,it still contains PHP Codes that can be evaluated by the Machine itself,however you need the markups,just use the file_get_contents,save to a variable,then you can append to innerHTML.that statement even if it works it bad coding.

You need to view the source code generated and see if there is an error.

The error i suspect is that, if the file.php contains a " it will break the script and generate a javascript error.

if that is the case (as i suspected it is), you need to change tactics. I must zip-dash outta here. . .

EasyPhp,

Can you put a sample page, so to examine it for you.