Build Dynamic Dependent Dropdowns in PHP Using AJAX - Real-World Example

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  • เผยแพร่เมื่อ 27 ม.ค. 2025

ความคิดเห็น • 8

  • @Codehow2
    @Codehow2  ปีที่แล้ว

    This tutorial is about creating a dynamic dropdown dependent on another. Watch this tutorial on how to create this type of dropdown using Ajax in PHP and MySQL. It will help you develop a similar application and enhance your knowledge of using Ajax in PHP. If you find it useful, please give it a like and subscribe to my channel.

  • @Oprekode
    @Oprekode ปีที่แล้ว +1

    K1 nice tutorial bro

    • @Codehow2
      @Codehow2  ปีที่แล้ว

      Thanks 🔥

  • @sKIBBSCHN
    @sKIBBSCHN 4 หลายเดือนก่อน

    Doesn't work for me. The $_POST variable won't be submitted and the result is always 0. If I change for $_POST to a fixed value all works fine. I can't find the error 😞.

    • @Codehow2
      @Codehow2  4 หลายเดือนก่อน

      Hi, just wanted to check if you downloaded the code from the website as given in the video descr. I just downloaded and ran, it worked fine for me. Which $_POST you are talking about? is it in get_item.php or index.php? Is Ajax working and form submission not happening? Let me know.

    • @sKIBBSCHN
      @sKIBBSCHN 4 หลายเดือนก่อน

      @@Codehow2 Good morning, first of all thank you for your answer. Since JavaScript is completely new to me, I rewrote the code while watching your video. That didn't work with $_POST (for you "categoty" for me "Country"). Was not transferred to the get_item.php. I then downloaded the code from the website. The JS file looks slightly different than in your video. I then adapted it to my form ID and it has worked ever since. Thanks for your answer and your work!

    • @Codehow2
      @Codehow2  4 หลายเดือนก่อน

      @@sKIBBSCHN glad to know that, thank you!

  • @redify1772
    @redify1772 ปีที่แล้ว +1

    The instructor has a very very nice tutorial ..people should subscribe snd learn a lot here