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Learn how to efficiently fetch and display data using AJAX with PHP, allowing you to show details in a dedicated div.
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This video is based on the question https://stackoverflow.com/q/63518412/ asked by the user 'Thillai Manalan Thallai Muthu' ( https://stackoverflow.com/u/13848478/ ) and on the answer https://stackoverflow.com/a/63518971/ provided by the user 'Haady Baabs' ( https://stackoverflow.com/u/7207967/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions.

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How to Display Ajax Response from Another Page in PHP by Utilizing the ShowDetails Div

When working with web applications, one common task is fetching data dynamically and displaying it on the webpage without needing to reload the page. This can be accomplished using AJAX (Asynchronous JavaScript and XML). However, a frequent conundrum developers face is: how do we display detailed information from one section of our application into another? In this guide, we will walk through how to utilize AJAX to achieve this, particularly focusing on populating the showDetails div when items from a list are clicked.

Understanding the Problem

You have a basic setup where clicking a button triggers an AJAX call that fetches data from the server and displays it in a list. This list is fetched based on user input and displayed in the showList div. However, you also want to display detailed information about each list item in the showDetails div when it's clicked. The original code does not have this feature implemented.

Current HTML Structure

In your index.php, the main HTML elements look like this:

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AJAX to Fetch Items

Your existing AJAX function looks like this:

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The Core Requirement

You need to implement functionality that allows users to click on an item from showList, which should trigger a new AJAX call to fetch details for that individual item and display it in the showDetails div.

Step-by-Step Solution

Step 1: Update Your List Items

Firstly, modify the way each item in the list is generated in the findsql.php. You should add an onclick event handler to each item that will pass the unique identifier (like number) into a new JavaScript function.

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Step 2: Create a New AJAX Function

In your index.php, create a new AJAX function (funcShowDetails). This function will take the unique number as an argument, making an AJAX call to fetch the detailed information from findsql.php.

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Step 3: Adjust Your Server-Side Script

In the findsql.php, you will need to add handling for the new AJAX call. Check if the id is set, and if so, modify the SQL query to fetch data related to that particular record.

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Conclusion

With these changes made, you can now click on items in your showList, and it will populate the showDetails div with the relevant information.

Wrapping Up

Using AJAX to dynamically display content allows for a more responsive and engaging user experience. By following the above steps, you've effectively set up a system for retrieving detailed records based on user interaction.

If you have any questions or need further clarification, feel free to comment below or reach out for help. Happy coding!