Posted by ljuljacka on 10/08/06 14:26

Yes, I get it now...tnx for your help. I guess I have some work to do :)
"ZabMilenko" <zabmilenko@hotmail.com> wrote in message
news:Sr7Wg.3$NT2.0@newsfe03.lga...
> Here is what I am seeing:
>
>
> while($filename=mysql_fetch_array($result))
> {
>
> for($i=0;$i<=count($filename);$i++)
> {
> // loop thru a bunch of files
>
> list($width, $height) = getimagesize($filename[$i]);
> // find out how big it is
>
> // Make an new image IN MEMORY
> $image_p = imagecreatetruecolor($new_width, $new_height);
>
> // make an image IN MEMORY from the original
> $image = imagecreatefromjpeg($filename[$i]);
>
>
> // Paint the image as a thumbnail IN MEMORY
> imagecopyresampled($image_p, $image, 0, 0, 0, 0, $new_width,
> $new_height, $width, $height);
>
>
> // Send the new image to the browser
> imagejpeg($image_p, null, 100)."\n";
> imagedestroy($image);
> imagedestroy($image_p);
> }
>
> // Repeat
> }
>
> The problem is you can only send one image per request. That is ONE image
> between header("content-type ...") and imagejpeg()
>
> You can make all you want, but only the first will go to the browser. The
> rest is as useless as spam.
>
> So you need to dump the thumbnails to disk and load them another way.
>
>
> ljuljacka wrote:
>> The thing I'm trying to get is to display images as thumbnails without
>> actually saving them, and the script does that but only with the first
>> image.
>> When I echo $filename[$i] inside the for loop it displays all file
>> locations.
>> When I removed imagedestroy() from script, it still does the same...
>> So the problem is displaying other images.
>>
>> If this answer is funny it's because I don't know if I understood you
>> correctly :)
>>
>> "ZabMilenko" wrote in message :
>>> It looks as if you are creating the second image, but not saving it
>>> anywhere.
>>>
>>> This generates an image: imagejpeg($image_p, null, 100)."\n";
>>>
>>> But what about $image? You do a copyresampled then destroy it. If you
>>> want to save the image, pass the second argument to imagejpeg.
>>>
>>>
>>

[Back to original message]