Table of Contents
Rendering Partial Views on Button Click in ASP.NET MVC
Introduction
Problem Overview
Solution
Code Example
Additional Considerations
Home Backend Development C++ How to Render Partial Views Dynamically on Button Click in ASP.NET MVC?

How to Render Partial Views Dynamically on Button Click in ASP.NET MVC?

Jan 03, 2025 pm 09:19 PM

How to Render Partial Views Dynamically on Button Click in ASP.NET MVC?

Rendering Partial Views on Button Click in ASP.NET MVC

Introduction

The task of displaying partial views dynamically on button click is a common requirement in ASP.NET MVC development. This article delves into a solution to this problem, demonstrating how to achieve this functionality using jQuery and the proper controller method.

Problem Overview

In ASP.NET MVC, partial views are typically rendered as a replacement for the entire page. However, in certain scenarios, it is desirable to render partial views within specific regions of the page only. For instance, when a user clicks a button, you may wish to load and display a partial view containing search results in a designated section of the page.

Solution

To accomplish this, follow the steps outlined below:

  1. Modify the Button: Replace the server-side button with an HTML button element (e.g., <button>).
  2. Implement jQuery Event Handler: Use jQuery's .click() event handler on the button to initiate the AJAX call.
  3. Pass Model Data: Send the necessary model data to the controller using jQuery's .load() method during the AJAX call.
  4. Receive Data in Controller: Define a controller method to receive the model data and return the appropriate partial view.
  5. Render Partial View: Update the designated page region with the partial view using jQuery's .load() method.

Code Example

Below is an example code implementation to render a partial view on button click:

Razor View (Index.cshtml)

<button>
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JavaScript

var url = '@Url.Action("DisplaySearchResults", "Search")';
$('#search').click(function() {
  var keyword = $('#Keyword').val();
  $('#searchResults').load(url, { searchText: keyword });
});
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Controller (SearchController)

public ActionResult DisplaySearchResults(string searchText)
{
  var model = // Build list based on the searchText
  return PartialView("SearchResults", model);
}
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Additional Considerations

  • Disable default form submission for buttons within forms using the type="button" attribute.
  • Use the @Html.ValidationMessageFor() helper only if the model is returned to the view.
  • Handle form submission with jQuery's .submit() event for models with validation attributes.

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