Thumbnail generation won't output anything

Question

I am trying to create simple thumbnail generation. I based it on another question here on Stack Overflow, but simplified the code for my needs. It's supposed to take an image and shrink it based only on height.

function create_thumbnail($original_pic, $intended_heigth){
$info = getimagesize($original_pic);
$actual_width = $info[0];
$actual_height = $info[1];

if($info['mime'] == 'image\jpeg'){
    $src = imagecreatefromjpeg($original_pic);
}else{
    return false;
}

$ratio = $intended_heigth / $actual_height;  
$newheight = $intended_heigth;
$newwidth = $actual_width * $ratio; 
$writex = 0;
$writey = 0;

$thumbnail = imagecreatetruecolor($newwidth, $newheight);
imagecopyresized($thumbnail, $src, $writex, $writey, 0, 0, $newwidth, $newheight, $actual_width, $actual_height);
return imagejpeg($thumbnail);
}

And then I am trying to echo id like this

<?php $original_pic = "images/info/7/01.jpg"; ?>
<img src="<?php create_thumbnail($original_pic, 90); ?>">

And this does nothing. But in the original code, there was the $writex defined this way $writex = round(($mintednded_width - $newwidth) / 2); But I don't really understand what is this even for. Any ideas?

http://stackoverflow.com/questions/4590441/php-thumbnail-image-resizing-with-proportions — rancor1223, Jul 12 '13 at 12:44
My knowledge of JS isn't sufficient for that and even though I doubt it would work. In the end I simply rewritten the way the thumbnails are created and implemented it to my CMS which now does it all for me. — rancor1223, Jul 15 '13 at 11:18

score 1 · Accepted Answer · answered Jul 12 '13 at 11:42

Your create_thumbnail function has the following return statement

return imagejpeg($thumbnail);

If you read the documentation for imagejpeg function, you will see that it returns a bool - whether image was created successfully or not.

And then you use that returned bool value for your <img>

<img src="<?php create_thumbnail($original_pic, 90); ?>">

What you want to do is return the path to which the generated thumbnail was saved. Read the documentation, pay attention to the second parameter of the imagejpeg function and use it to return path to saved thumbnail.

Good luck

Totally missed that. Thanks! – rancor1223 Jul 12 '13 at 19:11 — rancor1223, Jul 12 '13 at 19:11

score 0 · Answer 2 · answered Jul 12 '13 at 11:42

0

You incorrect use this function.

This function make thumbnail and return small image. You can store this image to file and then use link to this new image in your code.

You should change logic of your code.

answered Jul 12 '13 at 11:42

newman

2,689
15
23

score 0 · Answer 3 · answered Jul 12 '13 at 13:21

create image.php write bellow code
<?php
header('Content-Type: image/jpeg');

function create_thumbnail($original_pic, $intended_heigth) {
    $info = getimagesize($original_pic);
    $actual_width = $info[0];
    $actual_height = $info[1];

    if ($info['mime'] === 'image/jpeg') {
        $src = imagecreatefromjpeg($original_pic);
    } else {
        return false;
    }

    $ratio = $intended_heigth / $actual_height;
    $newheight = $intended_heigth;
    $newwidth = $actual_width * $ratio;
    $writex = 0;
    $writey = 0;

    $thumbnail = imagecreatetruecolor($newwidth, $newheight);
    imagecopyresized($thumbnail, $src, $writex, $writey, 0, 0, $newwidth, $newheight, $actual_width, $actual_height);
    return imagejpeg($thumbnail);
}
?>
<?php  $original_pic = "images/info/7/".$_GET['img']; ?>
<?php  create_thumbnail($original_pic, 90); ?>

Now you can call image with other file
<img src="image.php?img=01.jpg">

This is good, but it wouldn't work on my site as these thubnails are part of JS gallery so passing data like that isn't possible. But it's halpful anyway. Thanks. — rancor1223, Jul 12 '13 at 19:11

Thumbnail generation won't output anything

3 Answers3