pass php variable as argument in javascript function?

[xjasonx]

Heres two parts of the code:

 

echo "<script language=\"javascript\">";
echo "function openIt(x,y)";
echo "{";
echo "window.open(\"nwindow.php?id=arguments[0]&title=arguments[1]\",\"newwindow\",config=\"height=500,width=400\")";
echo "}";
echo "</script>";

 

"<a href=\"\" onClick=\"openIt($id,$title) \">Code</a></td>";

 

Then I get this error: Error:

missing ) after argument list
Source File: http://scriptsforgames.com/final_fantasy/scripts.php
Line: 1, Column: 17
Source Code:
openIt(1,Fishing V.1) 

 

Please help. I've tryed to fix this for so long.

[kenrbnsn]

You need to surround the arguments with quotes, since they are strings. I'm assuming the string is part of an echo statement:

<?php
echo "<a href=\"\" onClick=\"openIt('$id','$title') \">Code</a></td>";
?>

 

Ken

[xjasonx]

Wow, that was simple. So that got rid of the error, but now when I open the new window, it displays the variables as arguments[1].

 

echo "window.open(\"nwindow.php?id=arguments[0]&title=arguments[1]\",\"newwindow\",config=\"height=500,width=400\")";

 

Thanks a lot for the help. I won't be able to reply until tomorrow because I'm about to go to work.