08-18 04:15 阅读 144

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XMLHttpRequest post 传递多个参数及服务器端读取

一直没搞定XMLHttpRequest  post方法如何传递多种参数，比如同时读取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);
        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']);