我使用這個功能,創造由用戶上傳的圖像的縮略圖,我發現在這裏: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($dir);
}
此功能工作正常,不正是我希望它,但儘管如此,我仍然從中得到了錯誤。請參閱下面的錯誤:
Warning: opendir(images/008/01/0000288988r.jpg,images/008/01/0000288988r.jpg) [<a href='function.opendir'>function.opendir</a>]: The directory name is invalid. (code: 267)
Warning: opendir(images/008/01/0000288988r.jpg) [<a href='function.opendir'>function.opendir</a>]: failed to open dir: No error
Warning: readdir() expects parameter 1 to be resource, boolean given
的問題,我想,是我傳遞一個實際的文件,而不只是一個目錄中,將參數的函數。這是$pathtoimages
和$pathtothumbs
的情況。該功能應該搜索傳遞給它的目錄以查找擴展名爲.jpg
的所有圖像。但我只想在上傳時上傳的一張圖片上執行該功能。有沒有辦法編輯這個功能來允許這個?
在此先感謝
這沒有奏效。更多錯誤現在和現在縮略圖不再被創建:( – reubenCanowski 2013-03-21 15:50:17
哪些錯誤?將它們粘貼在這裏 – 2013-03-21 16:04:44
我想通了,再次感謝幫助和檢查 – reubenCanowski 2013-03-21 16:08:44