[Resolved] Help Resizing uploaded image

graceout

Greetings!

I am succesfully uploading an image via a web-form, but i would like to take that image, resize it, and place the new image in a different folder.

I am using the following code, and I have finallymanaged to not get any errors. But when I check, I do not see the new image in the second folder. (Yes, both folders have appropriate permissions.)

The first (uploaded) image is placed in ../pics. I want the altered image to be placed in ../picsth

Below is the code I am using. The image manipulation commands are all supported.

Thanks in advance for any help!

// function that resizes the image

function resizeImage($filename, $max_width, $max_height)
{
list($orig_width, $orig_height) = getimagesize($filename);

$width = $orig_width;
$height = $orig_height;

taller

if ($height > $max_height) {
$width = ($max_height / $height) * $width;
$height = $max_height;
}

wider

if ($width > $max_width) {
$height = ($max_width / $width) * $height;
$width = $max_width;
}

$image_p = imagecreatetruecolor($width, $height);

$image = imagecreatefromjpeg($filename);

imagecopyresampled($image_p, $image, 0, 0, 0, 0,
$width, $height, $orig_width, $orig_height);

return $image_p;
}

//If a file is submited with the form

if ($file_name) {

$path_to_file = "../pics/";
$location = $path_to_file.$file_name;
copy($file,$location);

//retrieve the newly uploaded image for resizing
$newfile="../pics/".$file_name;

resizeImage($newfile, 150, 150);

$newpath_to_file = "../picsth/";
$location = $newpath_to_file.$file_name;

// $image_p is supposedly the function return.

copy($image_p,$location);

}

////////////////////////////////////////////////////

I've tried mixing and matching just about everything. I even (somehow) got it to place the resized image in the original folder which resulted in two identical file names!

Help!

rachel2004

I glossed over your code. If you say you were able to get your function to work, but it isn't working in the the desired directory, try checking that you can write to the desired directory and then if you know that you can, echo out the value of the variables $location, $newfile, $newpath, etc. to see if they are holding the value that you think they should be holding.

bradgrafelman

$image_p is an image resource, not a filename nor a file.

What you need to do is use [man]imagejpeg/man to output the image resource to a file, then move that file using the copy() command.

For more info, read about the output of [man]imagecopyresampled/man.

EDIT: Use

 tags for PHP code in your posts.

graceout

I have adjusted my code to add the following:

// earlier function

$newfile="../pics/".$file_name;

$newpath_to_file = "../picsth/";
$location = $newpath_to_file.$file_name;
copy(imagejpeg(resizeImage($newfile, 150, 150)),$location);

I'm not sure how, but this code majically displays the proper thumbnail on the page! (Not what I want.)

It doesn't copy the new image file to the requested folder.

How can this be?

bradgrafelman

Did you read the manual page on [man]imagejpeg[/man] ? I think not!

Here's an example (untested):

$newpath_to_file = "../picsth/"; 
$location = $newpath_to_file.$file_name; 
$image = resizeImage($newfile, 150, 150);
imagejpeg($image, $location, 100);

graceout

Please don't be condescending. THis is new to me, and I DID read through that page.

It said the second and third fields were optional, and you told me to use copy() so I tried a variety of ways including the one I posted. I'm probably guilty of trying to stack too many functions together.

I apprecate you giving me the chance to work through this and learn. I am NOT looking to be spoon fed.

It also surprised me that it rendered rather than copying.

Thanks again for your patience and I'll post my sucess here for others.

Jon

bradgrafelman

Well... imagejpeg() can output the data to the buffer (which is what happens if you don't specify the 2nd argument); it does NOT return the image (meaning you can store it in variable).

Knowing this, I figured you could just supply the 2nd argue as the path where you want the file to go, and it'll just output it there.

I'm a bit confused on your response... did you say this solved it? Or was at least a step in the right direction? Either way, be sure to post new errors/problems/successes.

graceout

Thanks for your patience.

I was able to use the 'location' portion of imagejpg() to succesfully finish my project!

I appreaciate your teaching.

Thanks again,

Jon