Url Problem


i need to open a new window in my application in order to open a file edit it and then save it.

i am under the following (the script where i write the code is under this path) path:


the folder to opens that contains the file to open is under the Templates folder


in _form.php i need to write the code of a link or button that when i clicked, the folder window appeared.

i tried with open.window(files/Templates) or open.window(./files/Templates) or open.window(../../files/Templates) always the result: Unable to resolve the request "files".

how to write the correct url to open the folder?

any one can help?

thank you

protected is protected. You can’t access it from your browser.

you mean that if i need to access to the files/Templates folder, i have to move them out of the protected folder?

if yes of can i write its link?

thank you

i moved the files/Templates folder out the protected folder in this path:


i write this:

in config/main.php


                'homePath'=> dirname(__FILE__) . '/../../..',

		'protectedPath' => realpath($homePath . '/files'),


in _form.php

			$protectedPath = Yii::app()->params['protectedPath'];

			echo "<a href='$protectedPath'>link</a>";

			echo '<input type="button" value="New Window!" onClick="window.open(' . $protectedPath . ' ,\'Like\')">';

when press on the button in firefox and ie, an error appears:

SyntaxError: missing ) after argument list window.open(C:\xampp\htdocs\yii\basapp\files ,‘Like’) [after C:]