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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
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);
}The above is the detailed content of How to Dynamically Render a Partial View in ASP.NET MVC on Button Click?. For more information, please follow other related articles on the PHP Chinese website!
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