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Rendering Partial View Dynamically upon Button Click in ASP.NET MVC
Home Backend Development C++ How to Dynamically Render a Partial View in ASP.NET MVC on Button Click?

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

Jan 04, 2025 pm 10:01 PM

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

Rendering Partial View Dynamically upon Button Click in ASP.NET MVC

Problem Description:

In ASP.NET MVC, you want to render a partial view inside a specific div on the client side when a button is clicked, and the partial view should use data obtained dynamically during the button event.

Code Example:

Consider the following button code:

<button type="button">

This button attempts to navigate to a separate URL when clicked. To instead handle the click event and dynamically render the partial view, we need to modify the code as follows:

<button>

Additionally, add the following script to the page:

var url = '@Url.Action("DisplaySearchResults", "Search")';
$('#search').click(function() {
  var keyWord = $('#Keyword').val(); // Retrieve the user-entered search text
  $('#searchResults').load(url, { searchText: keyWord });
});

Modify the controller method to accept the search text as an argument:

public ActionResult DisplaySearchResults(string searchText)
{
  // Build list based on the provided searchText and return the partial view
  var model = // To be implemented
  return PartialView("SearchResults", model);
}

Explanation:

  • The jQuery .load method makes an AJAX call to the specified URL, passing any data provided in the data parameter.
  • The controller method is responsible for generating the partial view based on the dynamically provided data.
  • The updated content is loaded into the #searchResults div, dynamically rendering the partial view.

Note: If your model class contains multiple properties with validation attributes, replace the onclick handling with the following code and use a submit button instead:

$('form').submit(function() {
  if (!$(this).valid()) { return false; }
  var url = '@Url.Action("DisplaySearchResults", "Search")';
  var form = $(this).serialize();
  $('#searchResults').load(url, form);
  return false;

In this case, the controller method should handle the complete SearchCriterionModel object:

public ActionResult DisplaySearchResults(SearchCriterionModel criteria)
{
  var model = // Build list based on the properties of criteria and return the partial view
  return PartialView("SearchResults", model);
}

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