CODEWITHSUNDEEP
phpimage-processinggd
Hrishikesh -Rishi- Choudhari
Hrishikesh -Rishi- Choudhari

Reputation: 12427

imagejpeg() doesnt give the output properly in php

I want to upload an image from disk, resize it, and then upload it to Amazon S3.

However, I cant get the proper image output from imagejpeg().

heres my code:

$sourceUrl = $_FILES['path']['tmp_name'];       
$thumbWidth = '100';
$thumbid = uniqid();
$img = imagecreatefromjpeg($sourceUrl);
$width = imagesx( $img );
$height = imagesy( $img );

// calculate thumbnail size
$new_width = $thumbWidth;
$new_height = floor( $height * ( $thumbWidth / $width ) );

// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );

// copy and resize old image into new image
imagecopyresampled($tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height);

// output the image 
imagejpeg($tmp_img);

// upload thumbnail to s3
$s3->putObjectFile($tmp_img, "mybucket", $thumbid, S3::ACL_PUBLIC_READ);

Firebug gives me this error :

illegal character
[Break on this error] (�����JFIF���������>CREATOR: g...(using IJG JPEG v62), default quality\n

If I modify imagejpeg this way,

imagejpeg($tmp_img, 'abc.jpg');

then I get the same error. :(

Can i get some help here please ?

Upvotes: 2

Views: 3443

Answers (4)

Hrishikesh -Rishi- Choudhari
Hrishikesh -Rishi- Choudhari

Reputation: 12427

Well guys thank you very much again, I screwed around a bit more and combining that with your responses I got this working as follows :)

$thumbid .= ".jpg";
$img = imagecreatefromgif($sourceUrl);
$width = imagesx( $img );
$height = imagesy( $img );

// calculate thumbnail size
$new_width = $thumbWidth;
$new_height = floor( $height * ( $thumbWidth / $width ) );

// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );

// copy and resize old image into new image
imagecopyresampled($tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height);

$path = '/var/www/1.4/wwwroot/cdn/'.$thumbid;
// output the image 
if(imagegif($tmp_img, $path)){                  
    $thumblink = "";
    // upload thumbnail to s3
    if($s3->putObjectFile($path, "mybucket", $thumbid, S3::ACL_PUBLIC_READ)){
        $thumblink = "http://dtzhqabcdscm.cloudfront.net/".$thumbid;
        imagedestroy($tmp_img);
    }
    return $thumblink;
}

Upvotes: 0

Pete
Pete

Reputation: 1773

1) $tmp_img is a resource not a file. You probably need to save the image to disc and use that for putObjectFile

2) You probably need to tell S3 that the file you're uploading is of type image/jpeg

Upvotes: 0

Maerlyn
Maerlyn

Reputation: 34107

If you check the documentation of imagejpeg you can see it outputs the image, it means the way you call it it gets sent to the browser. You can get it to save to a file the second way you call it - by passing a filename in the second parameter.

Also, $tmp_img is an image resource, not a ready-to-use image file.

I don't know how your upload function works, but: if you need the file contents to upload, do it like this:

ob_start();
imagejpeg($tmp_image);
$image_contents = ob_get_clean();
$s3->putObjectFile($image_contents, "mybucket", $thumbid, S3::ACL_PUBLIC_READ);

if you need a filename to upload:

$filename = tempnam(sys_get_temp_dir(), "foo");
imagejpeg($tmp_image, $filename);
$s3->putObjectFile($filename, "mybucket", $thumbid, S3::ACL_PUBLIC_READ);

Upvotes: 3

Radek Suski
Radek Suski

Reputation: 1392

You have to define the header:

header('Content-type: image/jpeg');

Upvotes: 0

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