Re: [PHP-DB] One more GD ? -- filename

[Craig Hoffman]

Hi Jos,
I understand what your saying.  The problem is all the image names are 
listed in the DB (image.jpg, image1.jpg and so on)  and all images have 
titles attached.  For example,
Title:  Large Green Sea Turtle
Image: LargeGreenSeaTurtle.jpg

As you can see, the GD script randomly pulls photo names from the DB 
and shows them.  When I output the image with using the image tag I get 
a blank variable.

<img src='include/function_gd.php?route_photo='  '' height='x' 
width='x' align='center' id='image'>

Since the image name is empty, I can not match/query the title and 
image.  The image output may be showing a  Blue Sea Turtle but the 
title says, Green Sea Turtle. If I could just get the image name, I 
could run a query to match them.  Does this make sense?


On Apr 6, 2005, at 10:36 AM, Juffermans, Jos wrote:
Hi,
If I understand correctly, you're using
"include/function_gd.php?route_photo=$thumb" as image source which is 
the
script you pasted. If you use a script as image source, the script 
should
output the image data - not just store the image.

I'm not sure what you're trying to do but if you have an image object 
in php
(eg $thumb), you should output that image:

<?php
        $thumb = ImageCreateTrueColor($new_width, $new_height);
        $tmp_image = ImageCreateFromJPEG("../images/climbs/$route_photo");
        ImageCopyResampled($thumb, $tmp_image, 0, 0, 0, 0, $new_width,
$new_height, $width_orig, $height_orig);
        header("Content-type: image/jpeg");           // tell the browser
to expect a JPG
        imagejpeg($thumb, "", 100);                   // send the image to
the browser (filename="" => output instead of save)
        exit();
?>
However, you cannot use that to show multiple images from 1 script.
Jos
-----Original Message-----
From: Craig Hoffman [mailto:[EMAIL PROTECTED]
Sent: 06 April 2005 15:24
To: php-db@lists.php.net
Subject: [PHP-DB] One more GD ? -- filename
Thanks everyone for answering my other post.  However, I'm still having
trouble passing the image name from the GD script to the image tag.
I'm including the GD script below. Any help would be great. Thanks -
Craig
//GD script
<?php
        include ("db.php");
        $query = "SELECT route_photo FROM routes WHERE route_photo IS NOT
NULL
ORDER BY RAND() LIMIT 1";
        $result = mysql_query($query, $db) or die(mysql_error());
while($row = mysql_fetch_array($result)) {
        
        //Photo varibale
                $route_photo = $row["route_photo"];
                
        //Start GD
        //New Image resize
                $new_width = 125;
                $new_height = 225;
                
        //Content Type
                header("Content-type: image/jpeg");
                
        //Start new image resize        
                list($width_orig, $height_orig) =
getImageSize("../images/climbs/$route_photo");
                        
                if ($new_width && ($width_orig < $height_orig)) {
                                $new_width = ($new_height / $height_orig) *
$width_orig;
                        } else {
                                $new_height = ($new_width / $width_orig) *
$height_orig;
                                }
        
        //resample the image
                $thumb = ImageCreateTrueColor($new_width, $new_height);
                $tmp_image =
ImageCreateFromJPEG("../images/climbs/$route_photo");
                ImageCopyResampled($thumb, $tmp_image, 0, 0, 0, 0,
$new_width,
$new_height, $width_orig, $height_orig);
        
                ImageJPEG($thumb, $route_photo, '100');
                ImageDestroy($thumb);
        }
?>
//Image Tag:
echo ("<img src='include/function_gd.php?route_photo=$thumb'
align='center' id='image'>");
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