Image displaying

**ZabMilenko** · Oct 8 '06, 12:35 PM

Re: Image displaying

It looks as if you are creating the second image, but not saving it anywhere.

This generates an image: imagejpeg($imag e_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';
$result=mysql_q uery($query);
$new_width = 200;
$new_height = 150;
>
>
while($filename =mysql_fetch_ar ray($result)){
>
for($i=0;$i<=co unt($filename); $i++){
>
list($width, $height) = getimagesize($f ilename[$i]);
$image_p = imagecreatetrue color($new_widt h, $new_height);
$image = imagecreatefrom jpeg($filename[$i]);
imagecopyresamp led($image_p, $image, 0, 0, 0, 0, $new_width,
$new_height, $width, $height);
imagejpeg($imag e_p, null, 100)."\n";
imagedestroy($i mage);
imagedestroy($i mage_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><i mg 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
>
>

**ljuljacka** · Oct 8 '06, 01:05 PM

Re: Image displaying

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($imag e_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.
>
>

**ZabMilenko** · Oct 8 '06, 01:45 PM

Re: Image displaying

Here is what I am seeing:

while($filename =mysql_fetch_ar ray($result))
{

for($i=0;$i<=co unt($filename); $i++)
{
// loop thru a bunch of files

list($width, $height) = getimagesize($f ilename[$i]);
// find out how big it is

// Make an new image IN MEMORY
$image_p = imagecreatetrue color($new_widt h, $new_height);

// make an image IN MEMORY from the original
$image = imagecreatefrom jpeg($filename[$i]);

// Paint the image as a thumbnail IN MEMORY
imagecopyresamp led($image_p, $image, 0, 0, 0, 0, $new_width,
$new_height, $width, $height);

// Send the new image to the browser
imagejpeg($imag e_p, null, 100)."\n";
imagedestroy($i mage);
imagedestroy($i mage_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($imag e_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** · Oct 8 '06, 02:15 PM

Re: Image displaying

Yes, I get it now...tnx for your help. I guess I have some work to do :)
"ZabMilenko " <zabmilenko@hot mail.comwrote in message
news:Sr7Wg.3$NT 2.0@newsfe03.lg a...

Here is what I am seeing:
>
>
while($filename =mysql_fetch_ar ray($result))
{
>
for($i=0;$i<=co unt($filename); $i++)
{
// loop thru a bunch of files
>
list($width, $height) = getimagesize($f ilename[$i]);
// find out how big it is
>
// Make an new image IN MEMORY
$image_p = imagecreatetrue color($new_widt h, $new_height);
>
// make an image IN MEMORY from the original
$image = imagecreatefrom jpeg($filename[$i]);
>
>
// Paint the image as a thumbnail IN MEMORY
imagecopyresamp led($image_p, $image, 0, 0, 0, 0, $new_width,
$new_height, $width, $height);
>
>
// Send the new image to the browser
imagejpeg($imag e_p, null, 100)."\n";
imagedestroy($i mage);
imagedestroy($i mage_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($imag e_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.
>>>
>>>

>>

**Oli Charlesworth** · Oct 8 '06, 02:15 PM

Re: Image displaying

ZabMilenko said the following on 08/10/2006 14:55:

Here is what I am seeing:
>
>
while($filename =mysql_fetch_ar ray($result))
{
>
for($i=0;$i<=co unt($filename); $i++)
{
// loop thru a bunch of files
>
list($width, $height) = getimagesize($f ilename[$i]);
// find out how big it is
>
// Make an new image IN MEMORY
$image_p = imagecreatetrue color($new_widt h, $new_height);
>
// make an image IN MEMORY from the original
$image = imagecreatefrom jpeg($filename[$i]);
>
>
// Paint the image as a thumbnail IN MEMORY
imagecopyresamp led($image_p, $image, 0, 0, 0, 0, $new_width,
$new_height, $width, $height);
>
>
// Send the new image to the browser
imagejpeg($imag e_p, null, 100)."\n";
imagedestroy($i mage);
imagedestroy($i mage_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.

Or have the script only output one image, and call it N times, i.e.:

<img src="script.php ?img=1">
<img src="script.php ?img=2">
<img src="script.php ?img=3">
<img src="script.php ?img=4">

Image displaying

Image displaying

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