SEO Services
Website Programming in PHP / MYSQL
imagecreatefromjpeg
(PHP 4, PHP 5)
imagecreatefromjpeg — Create a new image from file or URL
Description
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
<?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:
A URL can be used 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.
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.
Windows versions of PHP prior to PHP 4.3.0 do not support access of remote files via this function, even if allow_url_fopen is enabled.
