ajax simulate form contain file to submit

fileupload.html

<script type="text/javascript">      
$(document).ready(function () {          
    $('#form1').submit(function () {              
    var formdata = new FormData();              
    var fileObj = document.getElementById("fileToUpload2").files;              
        for (var i = 0; i < fileObj.length; i++)                  
        formdata.append("file" + i, fileObj[i]);              
        $.ajax({                  
            type: 'POST',                  
            url: '/Home/Upload2',                  
            data: formdata,                  
            /**                  
                            *必须false才会自动加上正确的Content-Type                  
             */                 
            contentType: false,                  
            /**                  
            * 必须false才会避开jQuery对 formdata 的默认处理                  
            * XMLHttpRequest会对 formdata 进行正确的处理                  
            */                 
            processData: false             
        }).then(function () {                  
            alert('done');              
        }, function () {                  
            //failCal              
        });              
        return false;          
    });              
});         
function ajaxUpload() {          
    $("#form1").submit();      
}  
</script>