$path= $_SERVER['DOCUMENT_ROOT'] . "/img/".$HTTP_POST_FILES['fileupload']['name'];
if($ufile !=none)
{
if(copy($HTTP_POST_FILES['fileupload']['tmp_name'], $path))
{
echo "Successful<BR/>";
echo "File Name :".$HTTP_POST_FILES['ufile']['name']."<BR/>";
echo "File Size :".$HTTP_POST_FILES['ufile']['size']."<BR/>";
echo "File Type :".$HTTP_POST_FILES['ufile']['type']."<BR/>";
echo "<img src=\"$path\" width=\"150\" height=\"150\">";
}
else
{
echo "Error";
}
}
die();
为什么
echo "File Name :".$HTTP_POST_FILES['ufile']['name']."<BR/>";
不给我文件名?其他信息也没有得到回应
如果这是一个愚蠢的问题,我很抱歉,它与表单方法是 POST 有关吗?