Filter data using jQuery Ajax | Filter drop down list | PHP Tutorial

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  • Опубликовано: 7 янв 2021
  • Hi everyone, Here I had made a tutorial for you " Filter data using jQuery Ajax".
    Twitter: / geekwebsite
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Комментарии • 93

  • @ankandas
    @ankandas 2 года назад +4

    So far the best and easy explained video I ever watched for filters. Thank You so much..

  • @computerlearningbyargusaca5217
    @computerlearningbyargusaca5217 3 года назад +3

    🙏🙏Very nice . love to see more 👍👍videos . 😍😍 thanks for uploading👌👌

  • @user-zf4fs2kv8n
    @user-zf4fs2kv8n 11 месяцев назад

    Hi I am so happy after watching your video

  • @tarunjaiswal914
    @tarunjaiswal914 2 года назад

    Very flow and effectively explained thanks for making this vdo 👍👍👍👌👌

  • @firtoshkumar2115
    @firtoshkumar2115 2 года назад +2

    Awesome ji hadd khubsurat

  • @davidprincekachedwa1952
    @davidprincekachedwa1952 Год назад

    Thank you this is very helpful

  • @RicardoPires2023
    @RicardoPires2023 Год назад

    Thank you! It worked perfectly

    • @websitegeek7099
      @websitegeek7099  Год назад

      Thanks : )

    • @KonChEkey
      @KonChEkey Год назад

      got a source code u can share? cant get this to work for some reason

  • @slweb621
    @slweb621 2 года назад

    Thank you

  • @barankurt4316
    @barankurt4316 2 года назад +1

    Can you make a filter with province, district, neighborhood, categories? I want the cards suitable for filtering to be listed after the province, district, neighborhood categories are selected.

  • @martinknights5426
    @martinknights5426 2 года назад

    Well done and thank you for making the subject easy to understand the was you build each concept and then build on the concept is brilliant. Although I am not using Bootstrap of jQuery through watching your video i understand the principle and am confident I know what to build. Once again Thank You.

    • @websitegeek7099
      @websitegeek7099  2 года назад +1

      Thank You:) This means a lot to me.

    • @balkan77forum66
      @balkan77forum66 Год назад

      @@websitegeek7099 hi can you make it id student reigstration end date with msql?

  • @thierryjules6363
    @thierryjules6363 9 месяцев назад

    Hi
    Thanks for the wonderful video because i think it's a time eater.
    From my side,the filters doesn't work after a long weekend test.I made awrong but any issue.I had closely folowed 09:38
    possible to send my code please (where?)?

  • @JhonBenedictChannel
    @JhonBenedictChannel 2 года назад +1

    This code works for me. Thank you.

    • @nolep5555
      @nolep5555 2 года назад

      Can you share the code please?

  • @shah0308
    @shah0308 10 месяцев назад

    WebsiteGeek, Can you combine Ajax Filter drop down List and Livesearch on one PHP page?

  • @jibocobar8325
    @jibocobar8325 Год назад +1

    Thanks for the tutorial. It works on my project but the problem is I can only use the filter once. In order to use it again, I have to refresh the page. Can I request any suggestion? Thanks in advance.

    • @websitegeek7099
      @websitegeek7099  Год назад

      You can add the condition before $.ajax to check that if the var value is not empty then only our ajax code will run otherwise (else) page will remain as it is .. Basically we have to send the jQuery ajax request only if var value is not empty

  • @muhammedakram9378
    @muhammedakram9378 2 года назад

    What does this $(". container") code does, and is that derived from

    • @websitegeek7099
      @websitegeek7099  2 года назад

      Yes, $(".container") -- >This denotes that div section whose class is container (Here, )

  • @GyanGyan-ml6dw
    @GyanGyan-ml6dw Год назад

    My Table options filter table not work fetch.php not return data.

  • @jg9324
    @jg9324 2 года назад +1

    I like this . I will give it a try.

  • @nurulkhairisha1919
    @nurulkhairisha1919 Год назад

    Mine doesn't have error but my data not filter. I also already change the (;) to (,) Please help me

  • @sunandabhosle589
    @sunandabhosle589 3 года назад +1

    I have no error bt my data not filter after writing the code of fetch.php how can I do it

    • @websitegeek7099
      @websitegeek7099  3 года назад +1

      Make sure you passed the data parameter to the success function in index.php. Check this 09:38

  • @directorartstudio5448
    @directorartstudio5448 2 года назад +3

    how can i do it with multiple "filters"?

    • @mishelgonzalez4706
      @mishelgonzalez4706 6 месяцев назад

      Did you find how to do it? I'm trying to learn the same :(

  • @IbrahimAli-ph9nf
    @IbrahimAli-ph9nf 2 года назад

    i come with this error Uncaught TypeError: mysqli_num_rows(): Argument #1 ($result) must be of type mysqli_result, on fetchphp line 13 $count = mysqli_num_rows($result);

    • @websitegeek7099
      @websitegeek7099  2 года назад

      what do you have written in $query ?? Make sure you written the same code as in 7:52 .

  • @jandanielbacabac6282
    @jandanielbacabac6282 Год назад +1

    what version of ajax cdn does it support

  • @swathireddy1178
    @swathireddy1178 3 года назад +3

    i have an error in this place( data:'request=' + value;) Uncaught SyntaxError: Unexpected token ';'

    • @websitegeek7099
      @websitegeek7099  3 года назад

      You have to add comma (,) instead of (;) this semicolon

    • @carloskoa5751
      @carloskoa5751 3 года назад

      not sure if you guys cares but if you are bored like me during the covid times then you can stream all the new series on InstaFlixxer. I've been binge watching with my gf lately :)

    • @bobbyturner2477
      @bobbyturner2477 3 года назад

      @Carlos Koa Yup, been using InstaFlixxer for years myself =)

    • @felixespinosa7853
      @felixespinosa7853 2 года назад

      @@carloskoa5751 we don't care

  • @user-qy3ml3jp2h
    @user-qy3ml3jp2h Год назад +1

    I try to write ajax code but I did not get the pop alert this is the code
    $(document).ready(function(){
    $("#fetchval").on('change',function(){
    var value = $(this).val();
    alert(value);
    });
    });

  • @esanti10
    @esanti10 2 года назад +1

    Hi. I write this code but not working . It doesn't say where the error is. Can i send you my codes?

  • @sarasarsora5385
    @sarasarsora5385 Год назад

    How to creat that with javascript?

  • @wasss3745
    @wasss3745 Год назад

    hi can i get source code It didnt work to me

  • @user-qy3ml3jp2h
    @user-qy3ml3jp2h Год назад

    Please help me for this problem

  • @user-ir7ty1yy9z
    @user-ir7ty1yy9z 2 года назад

    Добрый день! Переписал код, фильтр не работает.
    Подскажите, где может быть ошибка.

  • @drofatfoed
    @drofatfoed 2 года назад +1

    What if I am to do this in WordPress, how will I go about it?

    • @websitegeek7099
      @websitegeek7099  Год назад

      You can use custom HTML & JS plugin for WordPress

  • @gumzhlayman5506
    @gumzhlayman5506 Год назад

    Dear sir, I have an issue with show display duplicate table when I selected the dropdown menu. How can I solve this issue?

  • @nolep5555
    @nolep5555 2 года назад

    Can you share the code please?

  • @advancewebdevelopment9802
    @advancewebdevelopment9802 2 года назад +1

    not workign

  • @aryananujsharma8683
    @aryananujsharma8683 Год назад

    How to do this simply with javascript in html and nothing else ?

    • @websitegeek7099
      @websitegeek7099  Год назад +1

      Instead of jQuery you can use javascript code also

  • @jepoydizon6766
    @jepoydizon6766 Год назад

    The work does not work for me

  • @ibrahimsaheed831
    @ibrahimsaheed831 2 года назад

    My output when it runs comes back as "working..."

    • @websitegeek7099
      @websitegeek7099  2 года назад

      Make sure you passed the data parameter to the success function in index.php. Check this 09:38..
      Check your jQuery ajax code again.

  • @binsarhrp9872
    @binsarhrp9872 Год назад +2

    Source code?

  • @user-ir7ty1yy9z
    @user-ir7ty1yy9z 2 года назад +1

    Спасибо за помощь! Like!!!

    • @nolep5555
      @nolep5555 2 года назад

      Can you share the code please?

    • @user-ir7ty1yy9z
      @user-ir7ty1yy9z 2 года назад

      @@nolep5555 Добрый день? Получилось скачать архив? Если нет, то дайте номер вайбера или почту. Я отправлю заново. А то здесь ссылки на FTP удаляют.

  • @karthi9750
    @karthi9750 Год назад

    in 4:51 after completing data page is not showing still in loading page

    • @websitegeek7099
      @websitegeek7099  Год назад

      Please check 9:36 , make sure you have inserted a comma after date option in ajax function.

  • @ashutosha372
    @ashutosha372 2 года назад +1

    I want the source code

  • @karthi9750
    @karthi9750 Год назад

    it show error in $r in line 57 why?

  • @namphuong4722
    @namphuong4722 Год назад

    hi i am facing 1 problem "Uncaught TypeError: $.ajax is not a function" can you help me how to fix it.
    thank you !!!

    • @websitegeek7099
      @websitegeek7099  Год назад

      Have you include the jQuery ajax CDN?

    • @namphuong4722
      @namphuong4722 Год назад

      @@websitegeek7099 yes i have followed all the steps

  • @greycatsdynasty1656
    @greycatsdynasty1656 5 месяцев назад

    Can you please share this code

  • @jokrisleti2540
    @jokrisleti2540 2 года назад

    I don't know why but $ajax is not working for me. And I am very sure that I followed all the steps.

    • @websitegeek7099
      @websitegeek7099  2 года назад

      Have you included the jQuery CDN under the tag ?

    • @nolep5555
      @nolep5555 2 года назад +1

      Can you share the code please?

  • @nitikanspl4614
    @nitikanspl4614 2 года назад

    my request is not wworking

    • @websitegeek7099
      @websitegeek7099  2 года назад +1

      You have to add comma (,) instead of (;) this semicolon in data option. Check on 09:39

  • @herbertbasilla3520
    @herbertbasilla3520 2 года назад

    why the variable request didn't work ?

    • @websitegeek7099
      @websitegeek7099  2 года назад

      Check your script 09:38 , make sure you have put comma after the data option
      [ data:'request=' + value, ]

    • @herbertbasilla3520
      @herbertbasilla3520 2 года назад +1

      @@websitegeek7099 thank you for your responce

    • @herbertbasilla3520
      @herbertbasilla3520 2 года назад +1

      @@websitegeek7099 sorry I got it, the problem is the class "container"

    • @nolep5555
      @nolep5555 2 года назад

      Can you share your code?. My code didn't working

  • @HosobaNaokiAChing
    @HosobaNaokiAChing 2 года назад

    what if 3 dropdown list bro?

    • @miloospins638
      @miloospins638 2 года назад

      Same problem, did you found an answer?

    • @HosobaNaokiAChing
      @HosobaNaokiAChing 2 года назад +1

      @@miloospins638 Yes bro, I changed the query in his code and I create my own query for 3 dropdown lists

  • @irhamrizky1965
    @irhamrizky1965 2 года назад

    Please send me source code ?

  • @ogungbuyivictor6278
    @ogungbuyivictor6278 Год назад

    source code

  • @Nightman-eb8mj
    @Nightman-eb8mj 4 месяца назад

    Disliked for not putting the source code...