Create a JPEG in PHP w/o creating a file?

Question

Here's my situation - I want to create a resized jpeg image from a user uploaded image, and then send it to S3 for storage, but am looking to avoid writing the resized jpeg to the disk and then reloading it for the S3 request.

Is there a way to do this completely in memory, with the image data jpeg formatted, saved in a variable?

Thanks again!

Answer 1

The Imagemagick library will let you do that. There are plenty of PHP wrappers like this one around for it (there's even example code for what you want to do on that page ;) )

Answer 2

Maye by using the GD library.

There is a function to copy out a part of an image and resize it. Of course the part could be the whole image, that way you would only resize it.

see imagecopyresampled

Answer 3

Once you've got the JPEG in memory (using ImageMagick, GD, or your graphic library of choice), you'll need to upload the object from memory to S3.

Many PHP S3 classes seem to only support file uploads, but the one at Undesigned seems to do what we're after here -

// Manipulate image - assume ImageMagick, so $im is image object
$im = new Imagick();
// Get image source data
$im->readimageblob($image_source);

// Upload an object from a resource (requires size):
$s3->putObject($s3->inputResource($im->getimageblob(), $im->getSize()), 
                  $bucketName, $uploadName, S3::ACL_PUBLIC_READ);

If you're using GD instead, you can use imagecreatefromstring to read an image in from a stream, but I'm not sure whether you can get the size of the resulting object, as required by s3->inputResource above - getimagesize returns the height, width, etc, but not the size of the image resource.

Answer 4

This can be done using the GD library and output buffering. I don't know how efficient this is compared with other methods, but it doesn't require explicit creation of files.

//$image contains the GD image resource you want to store

ob_start();
imagejpeg($image);
$jpeg_file_contents = ob_get_contents();
ob_end_clean();

//now send $jpeg_file_contents to S3

Answer 5

Most people using PHP choose either ImageMagick or Gd2

I've never used Imagemagick; the Gd2 method:

<?php

// assuming your uploaded file was 'userFileName'

if ( ! is_uploaded_file(validateFilePath($_FILES[$userFileName]['tmp_name'])) ) {
    trigger_error('not an uploaded file', E_USER_ERROR);
}
$srcImage = imagecreatefromjpeg( $_FILES[$userFileName]['tmp_name'] );

// Resize your image (copy from srcImage to dstImage)
imagecopyresampled($dstImage, $srcImage, 0, 0, 0, 0, RESIZED_IMAGE_WIDTH, RESIZED_IMAGE_HEIGHT, imagesx($srcImage), imagesy($srcImage));

// Storing your resized image in a variable
ob_start(); // start a new output buffer
  imagejpeg( $dstImage, NULL, JPEG_QUALITY);
  $resizedJpegData = ob_get_contents();
ob_end_clean(); // stop this output buffer

// free up unused memmory (if images are expected to be large)
unset($srcImage);
unset($dstImage);

// your resized jpeg data is now in $resizedJpegData
// Use your Undesigned method calls to store the data.

// (Many people want to send it as a Hex stream to the DB:)
$dbHandle->storeResizedImage( bin2hex($resizedJpegData) );
?>

Hope this helps.