Great video. I actually made it work on the first try! The only thing that would be nice is explanation on where the content goes if it's not inside a table.
Good job ! But today is Better to code with PDO. In fact, and you know that, the code can be swift between different system, and not only MySQL. Would you, for your next code to be published, do that with PDO. !!! Not only for me, but for all coder how are looking for some PHP code !!! Cheers...
Great tutorial, I would like to leave this suggestion for all who is going to implement this fantastic tutorial. Remove echo $sql at 29:25 and use $k = ( isset ( $_POST[ 'id' ] ) ? $_POST[ 'id' ] : '' ); at 24:09
Thank you so much for this great video. But how if I want to add information for the second mobile without removing the information for the first mobile? ... means I have to add a dynamic table row.. Can you help me with this please?information
Thank you very much for this tutorial, i have a question about if i want to pass the ajax response call data to another PHP file is it possible i tried sessions but it didn't work e.g $_SESSION[something] = $row[something]; (it didn't work) is there any other way to pass the data!!
Really helpful video. Is it possible to display the data in a text field instead of creating a table? I am looking to select a specific id from the database and display only the corresponding product name. Thanks in advance for any help.
It works great, but only one time. If I try to select from the dropdown and it crashes. If I refresh the page 2 or 3 times, it works. Is there a way to clear the cache after a selection?
Hi There, Thank you for the tutorial. I have used it in WordPress to develop a plugin and I am experiencing small issue with CORS: I am getting the following error Access to XMLHttpRequest at '' from origin ': Response to preflight request doesn't pass access control check: No 'Access-Control-Allow-Origin' header is present on the requested resource. When I use the CORS extension on my browser and disable it (CORS) works but only on my browser: Could you please help me resolve this issue.
hi sir had you succed to develop your plug in ? seems there is no such plug in (as i know ), if it is , could you share it please ? i need it, just answer me even you don;t want to share thank you
I tried this code but echo $sql the output is SELECT * FROM students WHERE sec_id = * I really need this kind of code for my capstone project can you help me? I can't display data 😰
Enable error reporting by adding the following code: ini_set('display_errors',1); Preferably at the start of the page. It will display the error, if any.
Awesome tutorial! I'm trying to do this with boxes though and I can't find a tutorial online _anywhere_ that shows how to do that. I'm looking and asking everywhere for help, so I'm asking for help with it here too. I asked on StackOverflow but some asshole commented "you're doing javascript, not database." and shut my question down (closed it) although there's ZERO javascript in the code I shared and I didn't type the word database even once in my question or title text. The question wasn't even posted for an hour and got zero answers or (helpful) comments. F%*K Stackoverflow!
Thanks for the tutorial. I had the dropdown list populating but now am getting an error related to the page you called showBrand. As you did I also echoed out the sql statement and it is breaking it because it does not recognise the $k variable. $k = $_POST['id']; // line 3 $k = trim($k); $con = mysqli_connect("localhost", "root", "password", "oct2020project"); $sql = "SELECT * FROM links WHERE subject='{$k}`"; echo $sql; The output from the echo is SELECT * FROM links WHERE subject='` This is happening because I am getting Notice: Undefined index: id in C:\xampp\htdocs\dependantdropdown2021\dependantdropdown\showlinks.php on line 3 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\dependantdropdown2021\dependantdropdown\showlinks.php on line 13 while($rows = mysqli_fetch_array($res)){ ?> //Line 13 I am using a table called links. Where you used brand I used subject.
Raj you are a genius sir. I'm working on a project and this was exactly what I was looking for.
Exactly what I was looking for. Thank you so much Raj!!!
Great work. It is very helpful and step by step explanation made it easy to understand. Thank you.
Thanks! This helped me to customize a crucial and difficult piece of my program. This is a great instructional video.
Great video. I actually made it work on the first try! The only thing that would be nice is explanation on where the content goes if it's not inside a table.
IT Finally Worksss !!!1 Thank YOU!!! ❤❤❤
Good job !
But today is Better to code with PDO. In fact, and you know that, the code can be swift between different system, and not only MySQL.
Would you, for your next code to be published, do that with PDO. !!! Not only for me, but for all coder how are looking for some PHP code !!!
Cheers...
Perfect tut, great work Big Raj
Thank you, sir. I appreciate slower pace so that also beginners can keep up with it.
Your actually the goat, This helped so much thank you.
What if I want to show more tables and drop downs on the side?
very useful becuse of this i resolve my issue very easly thank uhh
Waw, I really appreciated it! Thank you Raj Bhise
Great tutorial, I would like to leave this suggestion for all who is going to implement this fantastic tutorial. Remove echo $sql at 29:25 and use $k = ( isset ( $_POST[ 'id' ] ) ? $_POST[ 'id' ] : '' ); at 24:09
Thank you so much for this great video.
But how if I want to add information for the second mobile without removing the information for the first mobile? ... means I have to add a dynamic table row..
Can you help me with this please?information
thank you for the Video, but how I can add Select all to dropdown list?
Thank you
Thank you very much Brother
tried the same code but mine does not display results when searched
Thank you very much for this tutorial, i have a question about if i want to pass the ajax response call data to another PHP file is it possible i tried sessions but it didn't work e.g $_SESSION[something] = $row[something]; (it didn't work) is there any other way to pass the data!!
It would be good if you provide github repo link of such codes in the description of your videos.
Thankyousomuch sir!
really helpful brother.. god bless you.
if I were to add a filter for the model and show the data related to the filter, what should I do?
hi , thank you for this tutorial, i have just one question: is it posible to select one of the output rows and add it into a form?
Hello! Were you able to solve this problem? I am having the same one
Yes it is. Will share the link soon.
sir great video. Thank you sir
Really helpful video. Is it possible to display the data in a text field instead of creating a table? I am looking to select a specific id from the database and display only the corresponding product name. Thanks in advance for any help.
Yes you can!
ua-cam.com/video/29pVT2UoC6g/v-deo.html
Sir i want to fetch particular data under select tag which are stored in different tables...Please help me out
It works great, but only one time. If I try to select from the dropdown and it crashes. If I refresh the page 2 or 3 times, it works.
Is there a way to clear the cache after a selection?
Fantastic !!!!!!!
Amazing! I got it to work for me! THANKS!
how can we use database tables as options in dropdown menu to insert data in selected table....?
Thanks
Thank you so much, man! It helped me a lot😊
Bro it's not working ..
I write "echo $sql" in the end but the problem that happened to you stayed with me ...
Give me solution pleease
nice bro but can i contact with u cuz i have a problem in my code
How to do the same thing with a "Sidebar" or a "Navbar" instead of dropdown list..... Please Help !!
Hi There, Thank you for the tutorial. I have used it in WordPress to develop a plugin and I am experiencing small issue with CORS: I am getting the following error Access to XMLHttpRequest at '' from origin ': Response to preflight request doesn't pass access control check: No 'Access-Control-Allow-Origin' header is present on the requested resource.
When I use the CORS extension on my browser and disable it (CORS) works but only on my browser: Could you please help me resolve this issue.
hi sir had you succed to develop your plug in ? seems there is no such plug in (as i know ), if it is , could you share it please ? i need it, just answer me even you don;t want to share thank you
how do we edit the displayed data?
Perfect!
can you please show how to add pagination on this code
Hi, may i know why is twice? thanks in advance
how to show data in input field already on page ?
how to do same thing in laravel??
hi, nice job... source code?
I tried this code but
echo $sql the output is SELECT * FROM students WHERE sec_id = *
I really need this kind of code for my capstone project can you help me? I can't display data 😰
I am facing some problem can i share with you
yes you can.
Data not display without showing any error. How?
Enable error reporting by adding the following code:
ini_set('display_errors',1);
Preferably at the start of the page. It will display the error, if any.
I keep getting Failed to load resource: the server responded with a status of 500 () after selecting an option in the dropdown
check if you have installed php-mysql connector...
Bro it's not working for 2 selector combobox
can you clarify 2 selector combobox
@@rajbhise8586
I mean how to use 2 two drop-down list ??
Awesome tutorial! I'm trying to do this with boxes though and I can't find a tutorial online _anywhere_ that shows how to do that. I'm looking and asking everywhere for help, so I'm asking for help with it here too. I asked on StackOverflow but some asshole commented "you're doing javascript, not database." and shut my question down (closed it) although there's ZERO javascript in the code I shared and I didn't type the word database even once in my question or title text. The question wasn't even posted for an hour and got zero answers or (helpful) comments.
F%*K Stackoverflow!
ua-cam.com/video/29pVT2UoC6g/v-deo.html Check with this...it may be helpful...
Thanks for the tutorial. I had the dropdown list populating but now am getting an error related to the page you called showBrand. As you did I also echoed out the sql statement and it is breaking it because it does not recognise the $k variable.
$k = $_POST['id']; // line 3
$k = trim($k);
$con = mysqli_connect("localhost", "root", "password", "oct2020project");
$sql = "SELECT * FROM links WHERE subject='{$k}`";
echo $sql;
The output from the echo is
SELECT * FROM links WHERE subject='`
This is happening because I am getting
Notice: Undefined index: id in C:\xampp\htdocs\dependantdropdown2021\dependantdropdown\showlinks.php on line 3
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\dependantdropdown2021\dependantdropdown\showlinks.php on line 13
while($rows = mysqli_fetch_array($res)){ ?> //Line 13
I am using a table called links. Where you used brand I used subject.
PLEASE USE $id = (isset($_POST['id']) ? $_POST['id'] : '');
Urgent Urgent!!!!!- Excellent but i got an error msg "Warning: Undefined array key "id" in D:\xampp\htdocs\lesson\lesson1\showdata.php on line 2"
Same error ! Did you get any solution for the same ?
What if I want to show data with 5 drop-down list