
07-20-06, 02:13 AM
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$ problem...
Hi there..
in my php file this link is placed:
and this script to open a pop-up:
PHP Code:
echo " <SCRIPT LANGUAGE=\"JavaScript\">
var link
link = 'big.php?pic=$picture';
function pop4(locatie, hoogte12, breedte18) {
var breedte12 = 1024
var hoogte14 = 735
var winl = (screen.width-breedte12)/2;
var wint = (screen.height-hoogte14)/2;
var wint2 = screen.height-hoogte14;
window.open(link, \"\", \"height=400 , width=400 , top=50, left=50 ,scrollbars=auto, toolbar=0, resize=no\");
}
</SCRIPT> ";
the link in my php page looks like this:
unfortunatly $pic is empty in the file big.php.
The php in big .php looks like:
where did it go wrong?......
Last edited by jonnekke; 07-20-06 at 02:16 AM.
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07-20-06, 02:49 AM
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Quote:
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Originally Posted by jonnekke
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The variable is between single quotes, it won't be parsed. Plus, I'd use $_GET['pic'] instead of using extract().
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07-20-06, 02:54 AM
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that's wierd,
in another file on the same server (other folder)..
i use
and that's working perfect...
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07-20-06, 02:59 AM
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I tried you're solution...
but it's still nog working..

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07-20-06, 03:41 AM
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Look at the source code of big.php when you expect to see the image. What do you see there?
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07-20-06, 03:46 AM
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this is the source of big.php:
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07-20-06, 03:48 AM
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I mean when you see the page in your browser. Do rigth click in the window and hit "view page source".
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07-20-06, 03:52 AM
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07-20-06, 03:58 AM
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Ah,.. if your link looks like this
big.php expects to get a variable called "pic", not "picture". Try this.
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07-20-06, 04:02 AM
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my link looks like this:
and this is where is supposed to show the picture..
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