image gallery- image not displaying

[angelicdevil]

ok i have a folder made on the server and uploaded image files to it...the link to the image file or rather the path to the image is uploaded in the database in 'image_path' , now i want that the system read the path and resize it to generate a thumbnail and display it on the browser.

but the code is showing blank

 
<?
 
// generating a array with image paths 
 
$query_images = "SELECT image_path FROM userfolders" ;
$result_images = mysql_query($query_images);
confirm_query($result_images);
 
while ($record_images = mysql_fetch_assoc($result_images)) {
          $image_list[] = $record_images['image_path'] ;
 
?>
 
<?
// generating a thumbnail 
 
$thumb_height = 100;
 
for ($i=0,$i<count($image_list),$i+)
{
 
$filename = $image_list[$i];
list($width, $height) = getimagesize($filename);
$ratio = ($height/$width);
$newheight = $thumb_height;
$newwidth = ($thumb_height/$ratio);
 
$thumb = imagecreatetruecolor($newwidth, $newheight);
$source = imagecreatefromjpeg($filename);
 
$thumb_image = imagecopyresized($thumb, $source, 0, 0, 0, 0, $newwidth, $newheight, $width, $height);
 
  // displaying thumbnail
 
$thumbnail =  imagejpeg($thumb_image);
 
echo $thumbnail;
 
 
 
 
?>
 
 

Plz tell me wat i m doing wrong what lines i need to add or remove.

the image path is like d:/wamp/www/images/apple.jpg

[angelicdevil]

i took the static values as follwing code

<?php
 
// generating a array with image paths
 
 
 
// generating a thumbnail
$thumb_height = 100;
$filename = "http://bytes.com/images/apple.jpg";
list($width, $height) = getimagesize($filename);
        $ratio = ($height/$width);
        $newheight = $thumb_height;
        $newwidth = ($thumb_height/$ratio);
        echo  $newwidth;
        $thumb = imagecreatetruecolor($newwidth, $newheight);
        $source = imagecreatefromjpeg($filename);
 
        $thumbnail= imagecopyresampled($thumb, $source, 0, 0, 0, 0, $newwidth, $newheight, $width, $height);
 
// displaying thumbnail
 
header('Content-type: image/jpeg');
imagejpeg($thumb);
imagedestroy($thumb); 
 
 
 
 
 
?>
 
 

i get the output as

http://localhost/test.php

cant seem to figure out y as phpinfo shows gd is enabled

[Atli]

Two problems there...

1. To use URLs in that manner, the allow_url_fopen directive must be enabled.
2. The URL 'http://bytes.com/images/apple.jpg' isn't a image.

Try an absolute file system path instead.

Anyways, you need to debug the code... find out exactly where it is failing.
Check the return values of each of the function calls. Make sure they are actually valid. (See my first post.)

[angelicdevil]

code 1 :-

<?php
 
// generating a array with image paths
 
$query_images = "SELECT image_path,image_name FROM userfolders" ;
$result_images = mysql_query($query_images);
confirm_query($result_images);
 
while ($record_images = mysql_fetch_assoc($result_images)) {
          $file = $record_images['image_name'] ;
          $dirname = $record_images['image_path'] ;
 
 
// generating a thumbnail
$thumb_height = 100;
$filename = $dirname."/".$file;
list($width, $height) = getimagesize($filename);
        $ratio = ($height/$width);
        $newheight = $thumb_height;
        $newwidth = ($thumb_height/$ratio);
        echo  $newwidth;
 $thumb = imagecreatetruecolor($newwidth, $newheight);
 echo $thumb ;
 $source = imagecreatefromjpeg($filename);
 echo $source;
 $thumbnail= imagecopyresampled($thumb, $source, 0, 0, 0, 0, $newwidth, $newheight, $width, $height);
 echo $thumbnail;
 
 header('Content-type: image/jpeg');
imagejpeg($thumbnail);
 
}
?>
 

the o/p is as follows
109, resource id#10, resource id #12, 1

it seems the code is failing at imagejpeg command as soon as i put

code 2 :-

header('Content-type: image/jpeg');
imagejpeg($thumb);
imagedestroy($thumb); 
 

the o/p displays only http://localhost/test.php
even the values of new width thumb source and thumbnail do not get displayed when the code 2 is inserted into code 1

[Atli]

Your code is failing because of all the echos in your code.

If you plan to output an image, you must only output the image data, or the image will be corrupted and the browser will not be able to display it.

In cases like these, you should only print debug messages if there is a problem.Use if statements to check the variables and stop the execution with an error message if it doesn't check out.
(See the way I use die() in the example in my first post)

But don't print debug messages when outputting binary data unless there is a problem.

P.S.
You are passing the wrong variable to the imagejpg function in your first code.
It should be getting $thumb rather than $humbnail.