How to convert any image to JPG or JPEG using PHP

Question

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Save image from one format to another with php gd2

I am converting any image format to JPEG or JPG. Code for converting image to jpg is

function converttojpg($originalimg)
{
$filetype = @end(explode(".", $originalimg));
echo $originalimg . " asdasdfs";
if (strtolower($filetype) == 'jpg' or strtolower($filetype) == 'jpeg') 
    {
    $srcImg = imagecreatefromjpeg($originalimg);
    imagejpeg($srcImg,$originalimg,100);
    imagedestroy($srcImg);
    } 
if (strtolower($filetype) == 'png') 
    {
    $srcImg = imagecreatefrompng($originalimg);
    imagejpeg($srcImg,$originalimg,100);
    imagedestroy($srcImg);
    } 
if (strtolower($filetype) == 'gif') 
    {
    echo "executed";
    $srcImg = imagecreatefromgif($originalimg);
    imagejpeg($srcImg,$originalimg,100);
    imagedestroy($srcImg);
    }
}

After that i resize that converted image, whose code is

function resizeImage($originalImage,$toWidth,$toHeight)
{
// Get the original geometry and calculate scales
list($width, $height) = getimagesize($originalImage);
$xscale=$width/$toWidth;
$yscale=$height/$toHeight;

// Recalculate new size with default ratio
if ($yscale>$xscale)
    {
    $new_width = round($width * (1/$yscale));
    $new_height = round($height * (1/$yscale));
    }
else 
    {
    $new_width = round($width * (1/$xscale));
    $new_height = round($height * (1/$xscale));
    }
// Resize the original image
$imageResized = imagecreatetruecolor($new_width, $new_height);
$imageTmp     = imagecreatefromjpeg ($originalImage);
imagecopyresampled($imageResized, $imageTmp, 0, 0, 0, 0, $new_width, $new_height, $width, $height           );
return $imageResized;
}

and on Every image it reports errors and warning which are

    Warning: imagecreatefromjpeg() [function.imagecreatefromjpeg]: gd-jpeg, libjpeg: recoverable error: Corrupt JPEG data: 9 extraneous bytes before marker 0xd9 in E:\xampp\htdocs\photo\functions.php on line 33

Warning: imagecreatefromjpeg() [function.imagecreatefromjpeg]: 'userphotos/corrupted_image_1.jpg' is not a valid JPEG file in E:\xampp\htdocs\photo\functions.php on line 33

Warning: imagejpeg() expects parameter 1 to be resource, boolean given in E:\xampp\htdocs\photo\functions.php on line 34

Warning: imagedestroy() expects parameter 1 to be resource, boolean given in E:\xampp\htdocs\photo\functions.php on line 35

Warning: imagecreatefromjpeg() [function.imagecreatefromjpeg]: gd-jpeg, libjpeg: recoverable error: Corrupt JPEG data: 9 extraneous bytes before marker 0xd9 in E:\xampp\htdocs\photo\functions.php on line 22

Warning: imagecreatefromjpeg() [function.imagecreatefromjpeg]: 'userphotos/53.jpg' is not a valid JPEG file in E:\xampp\htdocs\photo\functions.php on line 22

Warning: imagecopyresampled() expects parameter 2 to be resource, boolean given in E:\xampp\htdocs\photo\functions.php on line 23
userphotos/MG_diesel-04.gif_thumb.png asdasdfsuserphotos/Porsche_911_996_Carrera_4S.jpg asdasdfs

Please help me in solving that problem. I have to complete it.

Not sure about gd extension. Try Imagick. http://php.net/manual/en/book.imagick.php — hjpotter92, Mar 18 '12 at 18:37
Using @ to suppress errors is never a good idea, especially if you are trying t work out why a script is not working as expected. — vascowhite, Mar 18 '12 at 19:37

score 0 · Answer 1 · answered Mar 18 '12 at 18:40

0

Don't use extension to determine file type. To check if given file is an image and in format that allows GD to manipulate it, use getimagesize

answered Mar 18 '12 at 18:40

dev-null-dweller

29,274
3
65
85

Can't agree. See [`Caution`](http://php.net/manual/en/function.getimagesize.php) section. – Boolean_Type Jul 09 '18 at 14:47
Of course for non-image files or unrecognized image format it won't be much helpful, but more common cases are jpeg files saved with png extension. – dev-null-dweller Jul 09 '18 at 21:45

score -1 · Answer 2 · answered Mar 18 '12 at 18:44

-1

<?php
    $image = @imagecreatefromstring(file_get_contents($file_path));
        if ($image === false) {
            throw new Exception (sprintf('invalid image or file not found, file=%s', $file_path));
        }
?>

imagecreatefromstring detects image type and creates an image resource, otherwise exception is thrown.

answered Mar 18 '12 at 18:44

Mikk

2,209
3
32
44

Using @ to suppress errors is never a good idea, especially if you are trying t work out why a script is not working as expected. – vascowhite Mar 18 '12 at 19:37

How to convert any image to JPG or JPEG using PHP

2 Answers2