Re: [PHP] Execute script then output image.

[JJ Harrison]

I found it.

I used header("Location: http://www.yoursite.com/images/stat.gif";);


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JJ Harrison
[EMAIL PROTECTED]
www.tececo.com

"David Otton" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> On Thu, 11 Jul 2002 22:01:03 +1000, you wrote:
>
> >> You need to output the correct content-type header for the image (eg
> >> image/gif), take it's size and output it as content-length, and
> >> suppress error reporting. Then output the image data.
>
> >There was a much simpler way. It just said the location of the file.
> >
> >I could use your method but the other one was much simpler.
>
> Is this what you meant?
>
> http://news.php.net/article.php?group=php.general&article=106348
> http://news.php.net/article.php?group=php.general&article=106370
>
> Looks like the same thing I suggested to me...
>
> $filename = "your/file.gif";
>
> // output the correct content-type header
> header("Content-Type: image/gif");
>
> // take it's size and output it as content-length
> header("Content-length:".filesize($filename));
>
> // Then output the image data
> readfile($filename);
>
> djo
>



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