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Home/ Questions/Q 9219689
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Editorial Team
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Editorial Team
Asked: June 18, 20262026-06-18T03:10:22+00:00 2026-06-18T03:10:22+00:00

I’m doing an hobby HTML project with lots of PHP, Jquery into it. I

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I’m doing an hobby HTML project with lots of PHP, Jquery into it.
I finally got my sorting working:
http://www.spisbilligt.nu/.

Now i need the picture to change when I have sorted the list and click one of the remaining buttons.

I got my code here:
http://jsfiddle.net/6qNAt/1/

 var itemInfo = { // Initialise the item information
  item1: ['item1.jpg', 'Description of item 1...'],
  item2: ['item2.jpg', 'Description of item 2...'],
  ...
  };

(function() {
  ('#items a').click(function() { // When an item is selected
        var id = $(this).attr('href').replace(/#/, ''); // Retrieve its ID
        ('#info img').attr('src', itemInfo[id][0]); // Set the image
        ('#info p').text(itemInfo[id][1]); // And text
        return false; // Prevent default behaviour
     });

});

To see what i am meaning and get hungry look here and see the cupcakes:
http://www.georgetowncupcake.com/menu.html

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-18T03:10:23+00:00Added an answer on June 18, 2026 at 3:10 am

    I would encourage you to move your data to data-* attributes on the list-items themselves:

    <!-- Storing data on the element is cleaner, and more manageable -->
    <ul id="items">
        <li data-src="pie_1.png" data-desc="Desc Item 1">Item 1</li>
        <li data-src="pie_2.png" data-desc="Desc Item 2">Item 2</li>
    </ul>
    

    Then use event-delegation to listen for any click:

    // Event Delegation is more efficient than many handlers
    $("#items").on("click", "li", function () {
    
    });
    

    And lastly, set the data attribute values to the image and text within the info area:

    $("#items").on("click", "li", function () {
        // Get the data from the clicked list item
        var data = $(this).data();
        // Use the data for our info elements
        $("#info")
            .find("img").attr("src", data.src).end()
            .find("p").html(data.desc);
    });
    

    Fiddle: http://jsfiddle.net/6qNAt/3/

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