XMLHttpRequest post 传递多个参数及服务器端读取
一贯没搞定XMLHttpRequest post方法怎样传递多种参数,比如一同读取post参数和file参数 var http = new XMLHttpRequest(); var form = new FormData(); // Add selected file to form form.append(me.getName(), file); http://www.fpzhangsha.com form.append('filename', '1.png'); // Send form with file using XMLHttpRequest POST request http.open('POST', me.getUrl()); http.send(form); 服务器端 $uploadfile = $uploaddir . basename($_FILES['userfile']['name']); $uploadfile = $uploaddir . basename($_POST['filename']); http://www.superkp123.com