Question

How do I use DevExtreme with Core MVC to send a model to the controller on form submit?

Answer and Explanation

To send a model to a controller using DevExtreme with Core MVC on form submit, you'll typically use DevExtreme's Form widget and configure it to serialize the form data into a model that can be received by your controller. Here's a breakdown of how to achieve this:

1. Set up your Core MVC Model:

- First, define the C# model that will represent the data you want to send from the form. For example:

public class MyModel
{
  public string Name { get; set; }
  public int Age { get; set; }
  public bool IsActive { get; set; }
}

2. Create a Controller Action:

- Create a controller action that will receive the model. Make sure it's decorated with the [HttpPost] attribute:

[HttpPost]
public IActionResult SubmitForm(MyModel model)
{
  if (ModelState.IsValid)
  {
    // Process the model data
    return Ok("Data received successfully!");
  }
  return BadRequest(ModelState);
}

3. Implement the DevExtreme Form:

- In your Razor view, use the DevExtreme Form widget. Ensure that the `formData` option is bound to a JavaScript object that matches the properties of your C# model. Also, set the `onContentReady` event to initialize the form and the `onSubmit` event to handle the form submission.

<div id="formContainer"></div>
<script>
  $(function() {
    $("#formContainer").dxForm({
      formData: {
        Name: "",
        Age: 0,
        IsActive: false
      },
      items: [
        { dataField: "Name", label: { text: "Name" } },
        { dataField: "Age", label: { text: "Age" }, editorType: "dxNumberBox" },
        { dataField: "IsActive", label: { text: "Is Active" }, editorType: "dxCheckBox" }
      ],
      onContentReady: function(e) {
        e.component.option("submitButtonOptions", {
          text: "Submit",
          useSubmitBehavior: true
        });
      },
      onSubmit: function(e) {
        var formData = e.component.option("formData");
        $.ajax({
          url: "/YourController/SubmitForm", // Replace with your controller action URL
          type: "POST",
          data: formData,
          success: function(result) {
            alert(result);
          },
          error: function(error) {
            alert("Error: " + JSON.stringify(error));
          }
        });
        e.preventDefault();
      }
    });
  });
</script>

4. Explanation:

- The `formData` option initializes the form with default values. The `items` array defines the form fields, mapping them to the properties of your model.

- The `onContentReady` event is used to customize the submit button.

- The `onSubmit` event handler prevents the default form submission and uses jQuery's `$.ajax` to send the form data to the controller. The `formData` is automatically serialized into a JSON object that matches the structure of your C# model.

5. Important Considerations:

- Ensure that the `url` in the `$.ajax` call matches the route to your controller action.

- Make sure you have included the necessary DevExtreme and jQuery libraries in your project.

- Handle errors and validation appropriately in both the client-side JavaScript and the server-side controller.

By following these steps, you can effectively use DevExtreme's Form widget to send a model to your Core MVC controller on form submit. This approach ensures that the data is correctly serialized and received by your server-side code.

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