Javascript中的PHP回显

Question

I have this code:

 document.getElementById('root').style.left = '<?php echo $page_position[$info["os"]][$info["browser"]][0]; ?>';
 document.getElementById('root').style.top = '<?php echo $page_position[$info["os"]][$info["browser"]][1]; ?>';

Why won't this work like this?

Can anybody point me in the right direction?

EDIT:

<?php echo $page_position[$info["os"]][$info["browser"]][1]; ?>

echoed "top:300px;"

Sorry guys, very stupid error of mine :/

Answer 1

如果将其放在.js文件中，则它将无法工作，因为它必须位于为PHP（ .php ）解析的页面中。

Answer 2

Things to check:

1. Check to make sure those variables are available. print_r($info) will help you out.
2. Make sure that the code is actually being filled with those variables (symptom of #1). View source and check that they are there.
3. Ensure you don't have any JavaScript errors preceding your code. Open your console (like Firebug) and check to make sure you don't.
4. Ensure that a root element exists. Again, use your console.

My money goes on #3, you probably have a JavaScript error somewhere that is stopping execution of the script.