Rendering Partial View on Button Click in ASP.NET MVC
Overview
This article addresses rendering a partial view on the client-side after a button click, resolving a common challenge in ASP.NET MVC applications. We will explore how to:
- Capture the button click event
- Send data to the controller
- Render the partial view inside a designated div
Problem Description
In a simplified ASP.NET MVC application, we seek to dynamically display search results in a div upon a button click. The challenge lies in fetching data via database access and then rendering the partial view using this data, without reloading the entire page.
Solution
1. Event Handling:
Replace the button with:
<button></button>
2. JavaScript:
Add the following script:
var url = '@Url.Action("DisplaySearchResults", "Search")'; $('#search').click(function() { var keyword = $('#Keyword').val(); $('#searchResults').load(url, { searchText: keyword }); })
- This script captures the click event and sends the search text to the controller using jQuery's .load() method.
3. Controller Method:
Modify the controller method to accept the search text:
public ActionResult DisplaySearchResults(string searchText) { var model = // Build model based on search text return PartialView("SearchResults", model); }
- This method fetches the data from the database and returns the partial view.
Additional Considerations
- Validation: If the SearchCriterionModel contains multiple properties with validation attributes, use a submit button and handle the form's .submit() event instead.
- Controller Method Signature: Update the controller method to receive the entire SearchCriterionModel object:
public ActionResult DisplaySearchResults(SearchCriterionModel criteria)
Tips for Rendering the Partial View Inside the Div
- Ensure the div has a unique ID, such as "searchResults".
- The partial view itself should have its own model class and data.
- The .load() method will update the contents of the div with the generated HTML from the partial view.
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