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.
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
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?)?
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.
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.
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 = $(this).val(); alert(value); }); });
not sure if you guys cares but if you are bored like me during the covid times then you can stream all the new series on InstaFlixxer. I've been binge watching with my gf lately :)
i come with this error Uncaught TypeError: mysqli_num_rows(): Argument #1 ($result) must be of type mysqli_result, on fetchphp line 13 $count = mysqli_num_rows($result);
So far the best and easy explained video I ever watched for filters. Thank You so much..
Thanks :)
Hi I am so happy after watching your video
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.
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
🙏🙏Very nice . love to see more 👍👍videos . 😍😍 thanks for uploading👌👌
Thank You!! :)
Awesome ji hadd khubsurat
Very flow and effectively explained thanks for making this vdo 👍👍👍👌👌
Thanks :)
WebsiteGeek, Can you combine Ajax Filter drop down List and Livesearch on one PHP page?
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?)?
Thank you this is very helpful
Thank you! It worked perfectly
Thanks : )
got a source code u can share? cant get this to work for some reason
what version of ajax cdn does it support
Ajax 3.5.1
how can i do it with multiple "filters"?
Did you find how to do it? I'm trying to learn the same :(
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.
can u give the source code? i didn't get any errors but the fikter doesn't work :(
Hi. I write this code but not working . It doesn't say where the error is. Can i send you my codes?
yes you can send me ..
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.
Thank You:) This means a lot to me.
@@websitegeek7099 hi can you make it id student reigstration end date with msql?
What if I am to do this in WordPress, how will I go about it?
You can use custom HTML & JS plugin for WordPress
My Table options filter table not work fetch.php not return data.
Source code?
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);
});
});
Add jQuery ajax CDN also
How to creat that with javascript?
This code works for me. Thank you.
Can you share the code please?
i have an error in this place( data:'request=' + value;) Uncaught SyntaxError: Unexpected token ';'
You have to add comma (,) instead of (;) this semicolon
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 :)
@Carlos Koa Yup, been using InstaFlixxer for years myself =)
@@carloskoa5751 we don't care
I have no error bt my data not filter after writing the code of fetch.php how can I do it
Make sure you passed the data parameter to the success function in index.php. Check this 09:38
I like this . I will give it a try.
Thanks :)
Mine doesn't have error but my data not filter. I also already change the (;) to (,) Please help me
I had the same issue.
did you fix it ?
Please help me for this problem
The work does not work for me
Dear sir, I have an issue with show display duplicate table when I selected the dropdown menu. How can I solve this issue?
You can send me your code
Hi! I also have the same problem, have you solved it now?
hi i am facing 1 problem "Uncaught TypeError: $.ajax is not a function" can you help me how to fix it.
thank you !!!
Have you include the jQuery ajax CDN?
@@websitegeek7099 yes i have followed all the steps
hi can i get source code It didnt work to me
How to do this simply with javascript in html and nothing else ?
Instead of jQuery you can use javascript code also
Добрый день! Переписал код, фильтр не работает.
Подскажите, где может быть ошибка.
What kind of error you are getting?
Thank you
What does this $(". container") code does, and is that derived from
Yes, $(".container") -- >This denotes that div section whose class is container (Here, )
not workign
I want the source code
Can you please share this code
in 4:51 after completing data page is not showing still in loading page
Please check 9:36 , make sure you have inserted a comma after date option in ajax function.
it show error in $r in line 57 why?
check your query or database connection file
why the variable request didn't work ?
Check your script 09:38 , make sure you have put comma after the data option
[ data:'request=' + value, ]
@@websitegeek7099 thank you for your responce
@@websitegeek7099 sorry I got it, the problem is the class "container"
Can you share your code?. My code didn't working
My output when it runs comes back as "working..."
Make sure you passed the data parameter to the success function in index.php. Check this 09:38..
Check your jQuery ajax code again.
Can you share the code please?
i come with this error Uncaught TypeError: mysqli_num_rows(): Argument #1 ($result) must be of type mysqli_result, on fetchphp line 13 $count = mysqli_num_rows($result);
what do you have written in $query ?? Make sure you written the same code as in 7:52 .
what if 3 dropdown list bro?
Same problem, did you found an answer?
@@miloospins638 Yes bro, I changed the query in his code and I create my own query for 3 dropdown lists
my request is not wworking
You have to add comma (,) instead of (;) this semicolon in data option. Check on 09:39
I don't know why but $ajax is not working for me. And I am very sure that I followed all the steps.
Have you included the jQuery CDN under the tag ?
Can you share the code please?
source code
Please send me source code ?
Спасибо за помощь! Like!!!
Can you share the code please?
@@nolep5555 Добрый день? Получилось скачать архив? Если нет, то дайте номер вайбера или почту. Я отправлю заново. А то здесь ссылки на FTP удаляют.
Disliked for not putting the source code...