(PHP 4, PHP 5)
imagecreatefromjpeg — Create a new image from file or URL
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);
?>
Ovenstående eksempel vil udskrive noget der ligner:
Du kan bruge en URL som filnavn med denne funktion, hvis fopen wrappers er slået til. Se fopen() for flere detaljer omkring hvordan man udformer filnavnet og List of Supported Protocols/Wrappers for en liste over understøttede URL-protokoller.
Path to the JPEG image
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 versioner af PHP før PHP 4.3.0 understøtter ikke adgang til fjernfiler via denne funktion, også selv om allow_url_fopen er slået til.