Thanks, but my html is valid, and what you are doing is not what I want to
do.
I want to display a random image included in the page, just like a banner
program would work.
I do not allow php execution on the page so the only way to get the image is
to call the script with the img src tag, just like a banner program would.
> First of all your HTML is not valid. Secondly I believe you're making it
> more complex than it needs to be. And since this is ending up in HTML
> anyway you don't need to concern yourself with the headers. Start off
with
> something simple like what I have below, and go from there.
>
> <html>
> <a href="show_image.php?image=1">Display myPicture.gif</a>
>
> <? // show_image.php script
> if ($image) {echo "<img src=\"http://www.myurl.com/mypicture.gif\";>;}
> ?>
> </html>
>
> Hope this helps,
> Kevin
>
> > Hi,
> > I am trying to do something like this:
> > <img src="image.php?image=1">
> >
> > The code is like this:
> >
> > <script language="php">
> > $szPicture = "myimage.gif";
> >
> > $url = "http://www.mysite.com/"; . $szPicture;
> > header("Content-type: image/gif");
> > header("Content-Length: " . strlen($url));
> > echo $url;
> > </script>
> >
> > But nothing happens, I just get the broken image.
> > My logs show nothing about a script failure or about a file not being
> found.
> > I tried without the header("Content-Length: " . strlen($url)); but no
> diff.
> >
> > Thanks for any help.
> >
> >
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![http://www.myurl.com/mypicture.gif\"](http://www.myurl.com/mypicture.gif\"){kind=link}