表单使用Ajax上传图片

HTML代码

<form id="up-form">
<input type="file" name="img" onchange="ajaxUp()" />
</form>

Javascript代码

function ajaxUp() {
      var data = new FormData($("#up-form")[0]);
      var url = "upload.php";
      $.ajax({  
        url: url,  
        type: 'POST',  
        data: data,  
        async: false,  
        cache: false,  
        contentType: false,  
        processData: false,  
        dataType: "json",  
        success: function (result) {  
          console.log(result.msg,false);
        },  
        error: function (result) {  
            console.log("网络传输错误",false); 
        }  
   });
}