Image resize PHP [duplicate]

Question

Possible Duplicate:
Can anybody suggest the best image resize script in php?

I'm still a newbie regarding image handling or file handling for that matter in PHP.

Would appreciate any input regarding the following

I post an image file using a simple html form and upload it via php. When i try and alter my code to accomodate larger files (i.e. resize) I get an error. Have been searching online but cant find anything really simple.

$size = getimagesize($_FILES['image']['tmp_name']);

//compare the size with the maxim size we defined and print error if bigger
if ($size == FALSE)
{
    $errors=1;
}else if($size[0] > 300){   //if width greater than 300px
    $aspectRatio = 300 / $size[0];
    $newWidth = round($aspectRatio * $size[0]);
    $newHeight = round($aspectRatio * $size[1]);
    $imgHolder = imagecreatetruecolor($newWidth,$newHeight);
}

$newname= ROOTPATH.LOCALDIR."/images/".$image_name; //image_name is generated

$copy = imagecopyresized($imgHolder, $_FILES['image']['tmp_name'], 0, 0, 0, 0, $newWidth, $newHeight, $size[0], $size[1]);
move_uploaded_file($copy, $newname); //where I want to move the file to the location of $newname

The error I get is:

imagecopyresized(): supplied argument is not a valid Image resource in

Thanks in advance

---

Thanks for all your input, i've changed it to this

$oldImage = imagecreatefromstring(file_get_contents($_FILES['image']['tmp_name']));
$copy = imagecopyresized($imgHolder, $oldImage, 0, 0, 0, 0, $newWidth, $newHeight, $size[0], $size[1]);
if(!move_uploaded_file($copy, $newname)){
    $errors=1;
}

Not getting a PHP log error but its not saving :(

Any ideas?

Thanks again

---

Result

Following works.

$oldImage = imagecreatefromjpeg($img);
$imageHolder = imagecreatetruecolor($newWidth, $newHeight);
imagecopyresized($imageHolder, $oldImage, 0, 0, 0, 0, $newWidth, $newHeight, $width, $height);
imagejpeg($imageHolder, $newname, 100);

Thanks for everyones help

Answer 1

imagecopyresized takes an image resource as its second parameter, not a file name. You'll need to load the file first. If you know the file type, you can use imagecreatefromFILETYPE to load it. For example, if it's a JPEG, use imagecreatefromjpeg and pass that the file name - this will return an image resource.

If you don't know the file type, all is not lost. You can read the file in as a string and use imagecreatefromstring (which detects file types automatically) to load it as follows:

$oldImage = imagecreatefromstring(file_get_contents($_FILES['image']['tmp_name']));