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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:
<code class="language-javascript">const formData = new FormData(document.querySelector('form'));</code>
This method efficiently handles any files uploaded via <input type="file">
elements.
AJAX POST Request
Send the FormData using AJAX:
<code class="language-javascript">$.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 } });</code>
Server-Side (Controller)
Within your controller, define the action to receive the data:
<code class="language-csharp">[HttpPost] public ActionResult YourActionName(YourModelType model) { // Process the model return View(); // Or any other appropriate return }</code>
Handling Files Separately (If Model Lacks HttpPostedFileBase
Property):
If your model doesn't include a property for HttpPostedFileBase
, handle file uploads separately:
<code class="language-csharp">[HttpPost] public ActionResult YourActionName(YourModelType model, HttpPostedFileBase uploadedFile) { // Process the model and uploadedFile return View(); }</code>
Adding Extra Properties:
To include properties not present in your form, append them to FormData:
<code class="language-javascript">formData.append('additionalProperty', 'additionalValue');</code>
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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