我想搜索文件並從jpg文件創建圖像。有代碼:搜索所有的文件和目錄在php中創建縮略圖
$dir2 = opendir($direction2);
$dir = opendir($direction);
function create_fcat($direction, $width, $direction2, $dir) {
while(false !== ($fn = readdir($dir))) {
$n_dir = $direction.'/'.$fn;
if (is_dir($n_dir) && $fn != '.' && $fn != '..') {
if ($handle = opendir($n_dir)) {
create_fcat($fn, $width, $direction2, $handle);
}
} elseif ($fn != '.' && $fn != '..') {
$ext = strtolower(substr($fn,strlen($fn)-3));
if ($ext == 'jpg') {
if ($img = imagecreatefromjpeg($direction.'/'.$fn)) {
$width_original = imagesx($img);
$height_original = imagesy($img);
if ($width_original > $height_original) {
$new_width = $width;
$new_height = floor($height_original * ($width/$width_original));
} else {
$new_height = $width;
$new_width = floor($width_original * ($width/$height_original));
}
$tmp_img = imagecreatetruecolor($new_width, $new_height);
imagecopyresized($tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width_original, $height_original);
imagejpeg($tmp_img, $direction2.'/large/'.$fn);
} else {
echo '<p>The file cannot be loaded.</p>';
}
}
}
}
closedir($dir);
}
的問題是:當例子,我們有以下文件:
- 主要
- image1.jpg
- FOLDER1
- image2.jpg
- 文件夾1。 1
- image3.jpg
文件image1.jpg和image2.jpg加載,但image3.jpg不是 - 它看起來像文件夾1.1被視爲文件。我該如何解決它?