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Posted by ZabMilenko on 10/08/06 12:51
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.
ljuljacka wrote:
> I'm trying to display resized images. Locations of images are fetched from
> database. The problem is that with the following code, I get only the first
> image displayed:
>
> [PHP]
> <?php
> header('Content-type: image/jpeg');
>
> $link = mysql_connect('localhost', 'root', '');
> if (!$link) {
> die('Could not connect: ' . mysql_error());
> }
>
> mysql_select_db("proba1");
> $query='select lokacija from galerija';
> $result=mysql_query($query);
> $new_width = 200;
> $new_height = 150;
>
>
> while($filename=mysql_fetch_array($result)){
>
> for($i=0;$i<=count($filename);$i++){
>
> list($width, $height) = getimagesize($filename[$i]);
> $image_p = imagecreatetruecolor($new_width, $new_height);
> $image = imagecreatefromjpeg($filename[$i]);
> imagecopyresampled($image_p, $image, 0, 0, 0, 0, $new_width,
> $new_height, $width, $height);
> imagejpeg($image_p, null, 100)."\n";
> imagedestroy($image);
> imagedestroy($image_p);
> }
> }
> [/PHP]
>
> With the next code in the same for loop I get all of the images, but
> ofcourse not resized:
>
> [PHP]
> ...
> echo "<table>";
> echo"<tr><td><img src=$filename[$i]></td></tr>";
> echo "</table>";
> ...
> [/PHP]
>
> So I guess the problem is in image functions, but I don't know how to solve
> it. Any suggestion is welcome :)
>
> Tnx,
> Dejan
>
>
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