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how JPG/JPEG to PNG and then PNG to JPG/JPEG

I need to convert user uploaded JPG/JPEG image to PNG and then PNG to back to JPG/JPEG.

How can I convert from JPG/JPEG to PNG and then PNG to JPG/JPEG?
0
php_beginner
Asked:
php_beginner
  • 9
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1 Solution
 
php_beginnerAuthor Commented:
1. i dont want to use image magick.
2. i need to convert the image using PHP, not some online tool. Will it be possible to convert the user uploaded images one by one use Online Image Converter? there will hundreds of images uploaded daily.
0
 
RoonaanCommented:
When your server supports it (most likely it does when it is hosted provider) you can use:

<?php
$im = createfromjpeg('myimage.jpg');
imagecreatepng($im, 'myimage.png');
?>
And
<?php
$im = createfrompng('myimage.jpg');
imagecreatejpeg($im, 'myimage.png');
?>

-r-
0
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php_beginnerAuthor Commented:
I am unable to find these functions in PHP.
0
 
php_beginnerAuthor Commented:
i have found a solution to convert PNG and then back to JPG. But there is an error on this line
$im = imagecreatefrompng($png_image);

The error is

<br />
<b>Warning</b>:  imagecreatefrompng(‰PNG

): failed to open stream: Invalid argument in <b>d:\apache\htdocs\test\gd.php</b> on line <b>22</b><br />
<br />
<b>Warning</b>:  imagejpeg(): supplied argument is not a valid Image resource in <b>d:\apache\htdocs\test\gd.php</b> on line <b>23</b><br />

Is it required to pass a physical image file resource?

<?
$image_source = "audio.jpg";

/********* PNG to JPG **********************/
ob_start();
header ("Content-type: image/png");
$im = imagecreatefromjpeg($image_source);
imagepng($im);
$png_image = ob_get_contents();
ob_end_clean();
/*******************************************/

/********* JPG to PNG **********************/
ob_start();
$quality = 99;
header ("Content-type: image/jpeg");
$im = imagecreatefrompng($png_image);
imagejpeg($im, '', $quality);
$jpg_image = ob_get_contents();
ob_end_clean();
/*******************************************/

?>

0
 
RoonaanCommented:
Yes, my fault. createfromjpeg/createfrompng had to be imagecreatefromjpeg/imagecreatefrompng.

The error you have in PNG to JPEG is because you haven't set a valid filename for the $image_source variable.

-r-
0
 
php_beginnerAuthor Commented:
>>The error you have in PNG to JPEG is because you haven't set a valid filename for the $image_source variable.

I didnt understand. Can you please correct that error?
0
 
RoonaanCommented:
You have to add a line:
$image_source = 'myimage.jpg'

Make sure that myimage.jpg is an actual image.

-r-
0
 
php_beginnerAuthor Commented:
I have this line on the top of the code. PNG image is creating correctly, when i echo $png_image it displays the image in the browser.

But giving error on $im = imagecreatefrompng($png_image);

<b>Warning</b>:  imagecreatefrompng(‰PNG

): failed to open stream: Invalid argument in <b>d:\apache\htdocs\test\gd.php</b> on line <b>22</b><br />
0
 
RoonaanCommented:
Please show full code. "PNG image is creating correctly" is somewhat obscure. $png_image just needs to be a filename/path to filename, nothing more, nothing less.

-r-
0
 
php_beginnerAuthor Commented:
I think you didnt read my posts carefully.

>>$png_image just needs to be a filename/path to filename, nothing more, nothing less.

In my previous post i asked "Is it required to pass a physical image file resource?".

This is the full source code and posting it 2nd time.

<?
$image_source = "audio.jpg";

/********* PNG to JPG **********************/
ob_start();
header ("Content-type: image/png");
$im = imagecreatefromjpeg($image_source);
imagepng($im);
$png_image = ob_get_contents();
ob_end_clean();
/*******************************************/

/********* JPG to PNG **********************/
ob_start();
$quality = 99;
header ("Content-type: image/jpeg");
$im = imagecreatefrompng($png_image);
imagejpeg($im, '', $quality);
$jpg_image = ob_get_contents();
ob_end_clean();
/*******************************************/

?>
0
 
RoonaanCommented:
You cannot use it like this. You indeed need to pass a physical string (as mentioned in the manual).
I am sorry for not reading correctly.

You can output to file using:
imagepng($im, 'image.png');

Then use $png_image = 'image.png';

-r-
0
 
php_beginnerAuthor Commented:
There should be an option to award points to myself :-)
0
 
php_beginnerAuthor Commented:
It means i need to save the image on disk and then use it. It will increase the IO operations:-s
0
 
RoonaanCommented:
You can also just keep using the $im variable, as that is the memory representation of an image, not specifically a JPG-image. Your above code can also be:

<?
$image_source = "audio.jpg";
ob_start();
header ("Content-type: image/png");
$im = imagecreatefromjpeg($image_source);
$quality = 99;
header ("Content-type: image/jpeg");
imagejpeg($im, '', $quality);
$jpg_image = ob_get_contents();
ob_end_clean();
/*******************************************/

?>

-r-
0
 
php_beginnerAuthor Commented:
In your code, i cant see you creating png image?
0
 
RoonaanCommented:
You are correct. Thought you where doing that just for the fun of it.
You can create png and jpg from same $im variable:
<?
$image_source = "audio.jpg";
ob_start();
header ("Content-type: image/png");
$im = imagecreatefromjpeg($image_source);
imagepng($im, 'audio.png');
$quality = 99;
header ("Content-type: image/jpeg");
imagejpeg($im, '', $quality);
$jpg_image = ob_get_contents();
ob_end_clean();
/*******************************************/

?>

-r-
0

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