How to append whole set of model to formdata and obtain it in MVC

How do I pass a whole set model object through formdata and convert it to model type in the controller?

Below is what I’ve tried!

JavaScript part:

model = {
             EventFromDate: fromDate,
             EventToDate: toDate,
             ImageUrl: imgUrl,
             HotNewsDesc: $("#txthtDescription").val().trim(),
        };
formdata.append("model",model);

then pass it through AJAX, it will be a string, and if I check the value of Request.Form["model"] the result will be same, that is it will be received as string and value will be "[object object]"

Is there any way to pass model through formdata and receive it in the controller?

If your view is based on a model and you have generated the controls inside <form> tags, then you can serialize the model to FormData using

var formdata = new FormData($('form').get(0));

This will also include any files generated with <input type="file" name="myImage" .../>

and post it back using

$.ajax({
  url: '@Url.Action("YourActionName", "YourControllerName")',
  type: 'POST',
  data: formdata,
  processData: false,
  contentType: false,         
});

and in your controller

[HttpPost]
public ActionResult YourActionName(YourModelType model)
{
}

or (if your model does not include a property for HttpPostedFileBase)

[HttpPost]
public ActionResult YourActionName(YourModelType model, HttpPostedFileBase myImage)
{
}

If you want to add additional information that is not in the form, then you can append it using

formdata.append('someProperty', 'SomeValue');

If you want to send Form data using Ajax.This is the way to send

var formData = new FormData();

//File Upload
   var totalFiles = document.getElementById("Iupload").files.length;


for (var i = 0; i < totalFiles; i++) {
    var file = document.getElementById("Iupload").files[i];

    formData.append("Document", file);
}

formData.append("NameCode", $('#SelecterID').val());
formData.append("AirLineCode", $('#SelecterID').val());


$.ajax({
        url: "/Controller/ActionName",
        type: "POST",
        dataType: "JSON",
        data: formData,
        contentType: false,
        processData: false,
        success: function (result) {
    }
})

Using Pure Javascript, considering you have

<form id="FileUploadForm">
   <input id="textInput" type="text" />
  <input id="fileInput" type="file" name="fileInput" multiple>
  <input type="submit" value="Upload file" />
</form>

JS

document.getElementById('FileUploadForm').onsubmit = function () {

var formdata = new FormData(); //FormData object

var fileInput = document.getElementById('fileInput');

//Iterating through each files selected in fileInput
for (i = 0; i < fileInput.files.length; i++) {
    //Appending each file to FormData object
    formdata.append(fileInput.files[i].name, fileInput.files[i]);
}
//text value
formdata.append("textvalue",document.getElementById("textInput").value);

//Creating an XMLHttpRequest and sending
var xhr = new XMLHttpRequest();
xhr.open('POST', '/Home/UploadFiles');
xhr.send(formdata); // se
xhr.onreadystatechange = function () {
    if (xhr.readyState == 4 && xhr.status == 200) {
        //on success alert response
        alert(xhr.responseText);
    }
  }
  return false;
}  

in your C# controller you can get values it as below

[HttpPost]
public ActionResult UploadFiles(YourModelType model, HttpPostedFileBase fileInput)
{
      //save data in db
}

Reference : File Uploading using jQuery Ajax or Javascript in MVC

In view side ,if you are using ajax then,

$('#button_Id').on('click', function(){
        var Datas=JSON.stringify($('form').serialize());
        $.ajax({
            type: "POST",
            contentType: "application/x-www-form-urlencoded; charset=utf-8",
            url: '@Url.Action("ActionName","ControllerName")',
            data:Datas,
            cache: false,
            dataType: 'JSON',
            async: true,
            success: function (data) {

            },
        });
    });

In Controller side,

 [HttpPost]
 public ActionResult ActionName(ModelName modelObj)
 {
 //Some code here
 }