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

Craig Hoffman Wed, 06 Apr 2005 12:02:49 -0700

Hi There (again), My apologies for being a complete pest with this issue. I changed a few things around and I can now pass the image name thanks to Jos, however, I'm getting a permission errors now.

Warning: imagejpeg(): Unable to open 'corneilus.jpg' for writing

I've changed image folder and subfolder permissions to 777 and no luck. Any ideas. Again thanks your help. :)

Craig

On Apr 6, 2005, at 11:18 AM, Juffermans, Jos wrote:

Hi,

Are you trying to display more images at once then?

You can store the image like you do now, then have the GD script output the image tag and use the image you've just stored as a source.

GD script:
...
ImageJPEG($thumb, $route_photo, '100');
echo '<img src="$route_photo" width="125" height="225" border="0"
title="$route_photo" alt="$route_photo" />';     // add this line
ImageDestroy($thumb);
...


in the other script instead of
        echo ("<img src='include/function_gd.php?route_photo=$thumb'
align='center' id='image'>");
do this:
        require("include/function_gd.php?route_photo=$thumb");

Jos

-----Original Message-----
From: Craig Hoffman [mailto:[EMAIL PROTECTED]
Sent: 06 April 2005 18:12
To: php-db@lists.php.net
Cc: Juffermans, Jos
Subject: Re: [PHP-DB] One more GD ? -- filename


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

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