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imagecreatefromjpeg

Create a new image from file or URL (PHP 4, PHP 5)
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 1001. Example to handle an error during creation

<?php
function LoadJpeg($imgname)
{
   
$im = @imagecreatefromjpeg($imgname); /* Attempt to open */
   
if (!$im) { /* See if it failed */
       
$im  = imagecreatetruecolor(150, 30); /* Create a black image */
       
$bgc = imagecolorallocate($im, 255, 255, 255);
       
$tc  = imagecolorallocate($im, 0, 0, 0);
       
imagefilledrectangle($im, 0, 0, 150, 30, $bgc);
       
/* Output an errmsg */
       
imagestring($im, 1, 5, 5, "Error loading $imgname", $tc);
   }
   return
$im;
}
header("Content-Type: image/jpeg");
$img = LoadJpeg("bogus.image");
imagejpeg($img);
?>

The above example will output something similar to:


Tip:

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 Appendix O, List of Supported Protocols/Wrappers for a list of supported URL protocols.

Parameters

filename

Path to the JPEG image

Return Values

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

Notes

Note:

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

Warning:

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.