Question

我有一个非常简单的照片上传器，需要一些提升速度

首先，当页面加载时，即使框中没有任何内容，也会出现回声？

if($_POST['upload']) {
     if($_FILES['image']['name'] == "")
     {
         #there's no file name return an error
         echo "\n<b>Please select a file to upload!\n</b>";
         exit;
     }
     #we have a filename, continue
}

#directory to upload to
$uploads = '/home/habbonow/public_html/other/quacked/photos';
$usruploads = 'photos';

#allowed file types
$type_array = array('image/gif','image/pjpeg','image/x-png');

if(!in_array($_FILES['image']['type'], $type_array))
{
    #the type of the file is not in the list we want to allow
    echo "\n<b>That file type is not allowed!\n</b>";
    exit;
}

页面输出显示上传框，但也回显“不允许使用该文件类型！”即使我没有点击按钮。

其次，jpg的mime类型是什么，因为我有jpeg和pjpeg。

谢谢，感谢任何帮助。

Answer 1

我还建议将所有内容放在POST块中，否则无论何时加载页面都会对其进行评估。

对于mimetypes，有一个方法image_type_to_mime_type，它允许你传入一个表示给定文件类型的常量，并为它返回正确的mimetype，例如：

$type_array = array(image_type_to_mime_type(), image_type_to_mime_type(IMAGETYPE_GIF), image_type_to_mime_type(IMAGETYPE_PNG), 'image/pjpeg');

（因为pjpeg没有它自己的常量，我们可以手动添加它）

Answer 2

如果您还没有提交表单，我认为！inarray调用很可能会返回false，因为$ _FILES ['images']不存在。

为此，我很想将所有内容放在第一个if语句中：

if($_POST['upload']) {
     if($_FILES['image']['name'] == "")
     {
         #there's no file name return an error
         echo "\n<b>Please select a file to upload!\n</b>";
         exit;
     }
     #we have a filename, continue


     #directory to upload to
     $uploads = '/home/habbonow/public_html/other/quacked/photos';
     $usruploads = 'photos';

     #allowed file types
     $type_array = array('image/gif','image/pjpeg','image/x-png');

     if(!in_array($_FILES['image']['type'], $type_array))
     {
         #the type of the file is not in the list we want to allow
         echo "\n<b>That file type is not allowed!\n</b>";
         exit;
     }
}

php照片上传器

2 个答案: