Jquery.data資料快取,儲存上傳的檔案,供儲存時一併提交
阿新 • • 發佈:2020-07-16
1 <div id="sto"> 2 <input type="file" onchange="upload(this)" multiple> 3 </div> 4 <script type="text/javascript"> 5 function upload(obj) { 6 var o = {}; 7 o["isa"] = true; 8 o["desc"] = "這是情況說明"; 9 o[前臺"files"] = obj.files; 10 $("#sto").data("myo", o); 11 $("#sto").empty(); 12 var f = $("#sto").data("myo"); 13 var formData = new FormData(); 14 formData.append("isa", f.isa); 15 formData.append("desc", f.desc); 16 for(var i = 0; i < f.files.length; i++) { 17 formData.append("upfile", f.files[i]); 18 } 19 $.ajax({ 20 url: "/Home/uploadfile", 21 type: "POST", 22 data: formData, 23 cache: false, //不設定快取 24 processData:false, // 不處理資料 25 contentType: false, // 不設定內容型別 26 success: function (data) { 27 alert("上傳了" + data.a + "個檔案。"); 28 } 29 }); 30 } 31 </script>
1 [HttpPost] 2 public ActionResult uploadfile(IEnumerable<HttpPostedFileBase> upfile,bool? isa,string desc) 3 { 4 return Json(new { a = upfile.Count()}); 5 }mvc後臺接收