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Скачать или смотреть How to Pass Variables from jQuery to PHP Using Ajax

  • vlogize
  • 2025-04-14
  • 7
How to Pass Variables from jQuery to PHP Using Ajax
passing variable from Jquery to php using ajaxjavascriptphpjqueryajax
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Описание к видео How to Pass Variables from jQuery to PHP Using Ajax

In this post, we delve into how to pass a variable from jQuery to PHP using Ajax while troubleshooting common pitfalls for a successful request.
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This video is based on the question https://stackoverflow.com/q/68277606/ asked by the user 'MrObscure' ( https://stackoverflow.com/u/12021185/ ) and on the answer https://stackoverflow.com/a/68277693/ provided by the user 'Dory Nguyen' ( https://stackoverflow.com/u/5808492/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions.

Visit these links for original content and any more details, such as alternate solutions, latest updates/developments on topic, comments, revision history etc. For example, the original title of the Question was: passing variable from Jquery to php using ajax

Also, Content (except music) licensed under CC BY-SA https://meta.stackexchange.com/help/l...
The original Question post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license, and the original Answer post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license.

If anything seems off to you, please feel free to write me at vlogize [AT] gmail [DOT] com.
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How to Pass Variables from jQuery to PHP Using Ajax: A Step-by-Step Guide

If you've ever encountered issues trying to send data from jQuery to PHP through an Ajax request, you're not alone. Many developers face challenges when dealing with Ajax, especially when the requests do not behave as expected. In this guide, we will explore a common problem: passing a jQuery variable to a PHP file using Ajax and how to troubleshoot it effectively.

Understanding the Problem

When you attempt to send a variable from jQuery to PHP using an Ajax POST request, the expected response might not come through. Below is the sample request code as an example of what may go wrong.

[[See Video to Reveal this Text or Code Snippet]]

Here's what your JavaScript function looks like:

[[See Video to Reveal this Text or Code Snippet]]

However, when running this code, you may encounter an error in your console:

[[See Video to Reveal this Text or Code Snippet]]

What Does This Error Mean?

This error message indicates that there is likely an issue with how jQuery is included in your project. Specifically, you might be using a version of jQuery that excludes Ajax functionality.

Solution Overview

1. Check Your jQuery Version
Ensure that you are using the full version of jQuery, not the slim version, which does not support Ajax. Here’s how to include the full version:

[[See Video to Reveal this Text or Code Snippet]]

2. Avoid Multiple Versions of jQuery
Make sure you do not have multiple instances of jQuery included in your file. Having different versions can lead to unexpected behavior.

3. Modify the Button Click Function
In your function logger, you are passing the button ID as a parameter (btni). However, you are not using it in your function. You might want to prevent the default button behavior to ensure that your Ajax call is triggered properly.

Update your function as follows:

[[See Video to Reveal this Text or Code Snippet]]

Sample PHP Code

Your PHP file handling the request might look like this:

[[See Video to Reveal this Text or Code Snippet]]

This basic script converts the incoming variable into a response that you can alert back in your JavaScript.

Conclusion

When dealing with Ajax requests, it’s crucial to ensure you are using the correct version of jQuery and check for multiple instances that might cause conflicts. Follow these steps, and you should be able to pass variables from jQuery to PHP successfully.

If you encounter further issues, consider revisiting your implementation, or feel free to drop a comment for assistance!

Stay curious and keep coding!

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