Front-end Tutorials

Front-end Tutorials

HTMLCSSSassJavaScriptReactJS
CMS Tutorials

CMS Tutorials

WordPress
Tutorials expand

jQuery Basic

jQuery Ajax Events

jQuery Ajax Methods

jQuery Keyboard Events

jQuery Keyboard Methods

jQuery Form Events

jQuery Form Methods

jQuery Mouse Event

jQuery Mouse Methods

jQuery Event Object

jQuery Fading

jQuery Document Loading

jQuery Traversing

jQuery Utilities

jQuery Property

jQuery HTML

jQuery CSS

jQuery Miscellaneous

jQuery ajaxSuccess Event

Posted in jQuery Tutorial
Updated on May 16, 2024
By Mari Selvan
👁️ 11 - Views
⏳ 4 mins
💬 0
jQuery ajaxSuccess Event

Photo Credit to CodeToFun

🙋 Introduction

In web development, handling AJAX (Asynchronous JavaScript and XML) requests is crucial for creating dynamic and interactive web applications. jQuery simplifies AJAX event handling with its ajaxSuccess event. However, it's important to note that .ajaxSuccess() has been deprecated in favor of .on().

In this guide, we'll explore the usage of the jQuery ajaxSuccess event, now implemented with .on(), providing examples to help you harness its power effectively.

🧠 Understanding ajaxSuccess Event

The ajaxSuccess event is triggered when an AJAX request completes successfully. It's particularly useful for executing code after a successful AJAX call, allowing you to update the UI or perform other actions in response to the data received.

💡 Syntax

The syntax for the ajaxSuccess event is straightforward:

syntax.js
Copied
Copy To Clipboard
.on("ajaxSuccess", [, eventData ], handler)

📝 Example

  1. Handling AJAX Success:

    Suppose you have an AJAX request that retrieves data from a server, and you want to display a success message when the request completes successfully. You can achieve this using the ajaxSuccess event:

    example.js
    Copied
    Copy To Clipboard
    $(document).on("ajaxSuccess", function(event, xhr, settings) {
      console.log("AJAX request successful!");
      console.log("Data received: " + xhr.responseText);
    });

    This will log a success message along with the received data to the console whenever an AJAX request succeeds.

  2. Updating UI After AJAX Success:

    You can also update the UI dynamically after a successful AJAX call using the ajaxSuccess event. For example, let's update the contents of a <div> element with the response data:

    index.html
    Copied
    Copy To Clipboard
    <div id="result"></div>
    example.js
    Copied
    Copy To Clipboard
    $(document).on("ajaxSuccess", function(event, xhr, settings) {
      $("#result").html(xhr.responseText);
    });

    This will replace the content of the result div with the data received from the AJAX request upon successful completion.

  3. Conditional Handling of AJAX Success:

    You can conditionally handle AJAX success events based on the request URL or other parameters. For instance, let's execute different actions for different AJAX requests:

    example.js
    Copied
    Copy To Clipboard
    $(document).on("ajaxSuccess", function(event, xhr, settings) {
      if (settings.url === "example.com/api/data") {
          console.log("Data retrieved successfully!");
      } else if (settings.url === "example.com/api/update") {
          console.log("Data updated successfully!");
      }
    });

    This will log specific messages based on the URL of the AJAX request.

🎉 Conclusion

The jQuery ajaxSuccess event, now implemented using .on(), provides a convenient way to handle AJAX success events in your web applications.

By leveraging this event, you can execute code after successful AJAX calls, update the UI dynamically, and even conditionally handle different AJAX requests. While .ajaxSuccess() is deprecated, transitioning to .on() ensures compatibility and continued support for your projects.

👨‍💻 Join our Community:

To get interesting news and instant updates on Front-End, Back-End, CMS and other Frameworks. Please Join the Telegram Channel:

Author

author
👋 Hey, I'm Mari Selvan

For over eight years, I worked as a full-stack web developer. Now, I have chosen my profession as a full-time blogger at codetofun.com.

Buy me a coffee to make codetofun.com free for everyone.

Buy me a Coffee

Share Your Findings to All

Subscribe
Notify of
guest
0 Comments
Inline Feedbacks
View all comments
We make use of cookies to improve our user experience. By using this website, you agree with our Cookies Policy
AgreeCookie Policy