How to save cropped image to a directory with Jcrop using PHP

Question

I can crop my images and click on crop. but then something goes wrong, because i don't know exactly how to save this image.

I use imagecreatefromjpeg from php. this code look like this:

<?php
SESSION_start();
if ($_SERVER['REQUEST_METHOD'] == 'POST')
{
$targ_w = 200;
$targ_h = 400;
$jpeg_quality = 90;

$src = $_SESSION['target_path'];
$img_r = imagecreatefromjpeg($src);
$dst_r = ImageCreateTrueColor( $targ_w, $targ_h );

imagecopyresampled($dst_r,$img_r,0,0,$_POST['x'],$_POST['y'],
$targ_w,$targ_h,$_POST['w'],$_POST['h']);

//header('Content-type: image/jpeg');
imagejpeg($dst_r, 'uploads/cropped/' . 'filename');

exit;
}

?>

My php code for saving the original image looks like this:

     <?php 
         session_start();
         $target = "uploads/"; 
         $target = $target . basename( $_FILES['filename']['name']) ; 
         $_SESSION['target_path'] = $target;

         $ok=1; 
         if(move_uploaded_file($_FILES['filename']['tmp_name'], $target)) 
         {
         echo "De afbeelding *". basename( $_FILES['filename']['name']). "* is geupload naar de map 'uploads'";
         } 
         else {
         echo "Sorry, er is een probleem met het uploaden van de afbeelding.";
         }
    ?> 

how do i save the cropped image? and is there a way to save the cropped image by overwriting the original so i get only the cropped image?

thanks in advance.

EDIT: I can save 1 photo now. i edited my code to: imagejpeg($dst_r, 'uploads/cropped/' . $filename .'boek.jpg'); But i have to make a function who can save multiple files with each another name. and maybe overwrite the original image, who is saved in 'uploads/'

Yalu

Does the folder you are trying to save to -- uploads\cropped -- exist? – Lacee

Then you should probably add error checking to all your image function calls. imagecreatefromjpeg, imagecreatetruecolor, imagecopyresampled, and imagejpeg all return false on failure. If you do something like if (! $img_r = imagecreatefromjpeg($src)) die ('here is the problem'); with all your calls you'll find the problem soon enough. – Lacee

but my code imagejpeg($dst_r, 'uploads/cropped/' . 'filename'); isn't right so hey uploads a file called filename now. but how do i call my image source so it uploads the image – Yalu

Any other people who know how to solve this? – Yalu

Answer 1

I am going to reply to what seems to be the question now:

But i have to make a function who can save multiple files with each another name. and maybe overwrite the original image, who is saved in 'uploads/'

An easy method to do this would be to add the timestamp in the filename, this way everytime the operation is executed, you get a new filename. Doing something along the lines of:

imagejpeg($dst_r, 'uploads/cropped/' . $filename . time() . 'boek.jpg');

Will save a new image everytime.

When you want to overwrite the original, simply assign the right path

imagejpeg($dst_r, 'uploads/' . $filename .'.jpg');

Edit: What seems to be the problem is getting the right file name, without hardcoding 'boek.jpg'.

The original code posted used the string filename when saving the image

imagejpeg($dst_r, 'uploads/cropped/' . 'filename');

What you want to do, is deduce the filename from $src. To do so, simply use the function basename (documentation here), so you will have:

imagejpeg($dst_r, 'uploads/cropped/' . basename($src)); // basename will already contain the extension

Baiss

I will try this. it looks so easy :) – Yalu

it saves the right image but with a name like this: 1354113593boek i actually want to have the original file called: boek1 overwrited to the new one. so i get no duplicated files. – Yalu

I guess I didn't understood what you meant by save multiple files with each another name. Let me rewrite this a bit... – Baiss

This is exactly what i mean :D it works as a charm now! Thankyou! – Yalu