How to Pass a Whole Model via FormData in MVC?

Submitting a Full Model via FormData in ASP.NET MVC

Transmitting a complete model object using FormData in ASP.NET MVC and subsequently deserializing it within the controller can be complex. This guide offers a straightforward solution.

Client-Side (JavaScript)

To convert your model into FormData, use:

const formData = new FormData(document.querySelector('form'));

This method efficiently handles any files uploaded via <input type="file"> elements.

AJAX POST Request

Send the FormData using AJAX:

$.ajax({
  url: '@Url.Action("YourActionName", "YourControllerName")',
  type: 'POST',
  data: formData,
  processData: false,
  contentType: false,
  success: function(response) {
    // Handle successful response
  },
  error: function(xhr, status, error) {
    // Handle errors
  }
});

Server-Side (Controller)

Within your controller, define the action to receive the data:

[HttpPost]
public ActionResult YourActionName(YourModelType model)
{
  // Process the model
  return View(); // Or any other appropriate return
}

Handling Files Separately (If Model Lacks HttpPostedFileBase Property):

If your model doesn't include a property for HttpPostedFileBase, handle file uploads separately:

[HttpPost]
public ActionResult YourActionName(YourModelType model, HttpPostedFileBase uploadedFile)
{
  // Process the model and uploadedFile
  return View();
}

Adding Extra Properties:

To include properties not present in your form, append them to FormData:

formData.append('additionalProperty', 'additionalValue');

This comprehensive approach simplifies the process of submitting and processing complete models using FormData in ASP.NET MVC. Remember to handle potential errors appropriately within the AJAX error callback.

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