%20 in my variable

Anon

Hi, I am just trying to display an image in a popup window (to show a larger image when the user clicks on the thumbnail) but for some reason the image name is being passed with a %20 (a space) in front of it, which i guess is why it cant find it....why is it doing this?....thanks for your help

<head>
<title><?php echo $title ?></title>
<body bgcolor=\"d4d4c3\">
<img src=\"images/<?php echo $image ?>\" border=\"0\">
</body>

the title works fine.

Anon

Try to use urlencode/urldecode functions to make it work around. What i do for opening a pop-up with a big image, wich is what you want to do, is the following:

<SCRIPT LANGUAGE="JavaScript">\n
function Zoom(image){
w=window.open(image,"Image","width=600,height=550,status=no,scrollbars=no");
}
</SCRIPT>

Anon

Not entirely sure how to use urlencode/urldecode, this is the function that i use and just the variables $title and $image, is this that different to what u do?

newWin = null

function opensmall(url) {
if (newWin && newWin.open && !newWin.closed) newWin.close();{
newWin = window.open(url,'newWin','width=400,height=300,top=200,left=100,scrollbars=no,menubar=no,resizable=no,taskbar=no,status=no')
}
}

Anon

silly me, it was passing a %20 cause i had a space between the equals and pulling out the variable silly me ...$image= $myrow[image_name]