我使用这个功能,创造由用户上传的图像的缩略图,我发现在这里:http://webcheatsheet.com/php/create_thumbnail_images.php:PHP createThumbnail功能故障排除
function createThumbs($pathToImages, $pathToThumbs, $thumbWidth)
{
// open the directory
$dir = opendir($pathToImages);
// loop through it, looking for any/all JPG files:
if (false !== ($fname = readdir($dir))) {
// parse path for the extension
$info = pathinfo($pathToImages . $fname);
// continue only if this is a JPEG image
if (strtolower($info['extension']) == 'jpg')
{
echo "Creating thumbnail for {$fname} <br />";
// load image and get image size
$img = imagecreatefromjpeg("{$pathToImages}{$fname}");
$width = imagesx($img);
$height = imagesy($img);
// calculate thumbnail size
$new_width = $thumbWidth;
$new_height = floor($height * ($thumbWidth/$width));
// create a new temporary image
$tmp_img = imagecreatetruecolor($new_width, $new_height);
// copy and resize old image into new image
imagecopyresampled($tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
// save thumbnail into a file
imagejpeg($tmp_img, "{$pathToThumbs}{$fname}");
}
}
// close the directory
closedir($dir);
}
此功能工作正常,不正是我希望它,但尽管如此,我仍然从中得到了错误。请参阅下面的错误:
Warning: opendir(images/008/01/0000288988r.jpg,images/008/01/0000288988r.jpg) [<a href='function.opendir'>function.opendir</a>]: The directory name is invalid. (code: 267)
Warning: opendir(images/008/01/0000288988r.jpg) [<a href='function.opendir'>function.opendir</a>]: failed to open dir: No error
Warning: readdir() expects parameter 1 to be resource, boolean given
的问题,我想,是我传递一个实际的文件,而不只是一个目录中,将参数的函数。这是$pathtoimages
和$pathtothumbs
的情况。该功能应该搜索传递给它的目录以查找扩展名为.jpg
的所有图像。但我只想在上传时上传的一张图片上执行该功能。有没有办法编辑这个功能来允许这个?
在此先感谢
这没有奏效。更多错误现在和现在缩略图不再被创建:( – reubenCanowski 2013-03-21 15:50:17
哪些错误?将它们粘贴在这里 – 2013-03-21 16:04:44
我想通了,再次感谢帮助和检查 – reubenCanowski 2013-03-21 16:08:44