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(PHP 4, PHP 5)

imagecreatefromjpeg — Create a new image from file or URL


resource imagecreatefromjpeg ( string $filename )

imagecreatefromjpeg() returns an image identifier representing the image obtained from the given filename.

On failure imagecreatefromjpeg() outputs an error message, which unfortunately displays as a broken link in a browser. To ease debugging the following example will produce an error JPEG:

Example#1 Example to handle an error during creation

function LoadJpeg($imgname)
$im = @imagecreatefromjpeg($imgname); /* Attempt to open */
if (!$im) { /* See if it failed */
$im  imagecreatetruecolor(15030); /* Create a black image */
$bgc imagecolorallocate($im255255255);
$tc  imagecolorallocate($im000);
/* Output an errmsg */
imagestring($im155"Error loading $imgname"$tc);
header("Content-Type: image/jpeg");
$img LoadJpeg("bogus.image");

The above example will output something similar to:


You can use a URL as a filename with this function if the fopen wrappers have been enabled. See fopen() for more details on how to specify the filename and List of Supported Protocols/Wrappers for a list of supported URL protocols.



Path to the JPEG image

Return Values

Returns an image resource identifier on success, FALSE on errors.


Note: JPEG support is only available if PHP was compiled against GD-1.8 or later.


Windows versions of PHP prior to PHP 4.3.0 do not support accessing remote files via this function, even if allow_url_fopen is enabled.