There is a code that produces photos added to the database. I tried to insert this code into regular js sliders, but all I got was a bunch of photos superimposed on each other and a slider that didn’t work. How can I make a slider out of this code? Here is how the code looks when this script is
Tag: jquery
Convert to accordion
I’m new here. I’m trying to convert my <li> into an accordion in the mobile view. I actually have something similar to: And I have this on the footer.php But it shows me “Uncaught TypeError: oTab is undefined” So every time I click on the li it displays the related div description. Now I’d like to convert that into an
Jquery – php event happening twice
I’ve got a simple chrome extension that is using Jquery and works on a page on our local network. When that page is loaded jquery reacts to clicks on an element called #site and then calls a php page. This has been working for the last 6 months without issue. Normally the test.php page is run once and looking at
The “PUT method is not supported for this route” error comes up after jquery ajax update
I update the database with jquery ajax and print “success/failure” with alert according to the return value. The product update is successful, but the message “failed” appears on the screen and I get the error “The PUT method is not supported for this route”. My jquery: url: coming here by route: Answer Change the method to POST. Considering that csrf
Fetch specific rows from MySQL with datatables
I’m trying to fetch specific rows in a MySQL table with specific “basket_id”s and display them in the jQuery DataTables plugin. In the PHP server-side script for the DataTables AJAX request, I specify the $user->basketId() in a WHERE clause for MySQL.. Let me walk you through it.. User signs in. USER class stores user session data. (use public methods to
PHP unable to identify sqlsrv_connect() function while trying to connect SQL Server
I’m trying to connect to my local SQL Server from a simple PHP file using the sqlsrv_connect() function, but every time I’m calling the file in the browser through localhost, it’s throwing a 500 (Internal Server Error) saying: PHP Fatal error: Uncaught Error: Call to undefined function sqlsrv_connect() in C:inetpubwwwrootAJAX_Tutorialget_db_data.php:4 get_db_data.php is the file from which I’m trying to connect
Show/hide multiple forms depending on dropdown without conditional statements [closed]
Closed. This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 5 months ago. Improve this question I need to show/hide three forms depending on dropdown selected option without using conditional statements. Is it possible? Answer You can do it like so,
how to display a footer button only on certain pages?
I have a page called footer.php which is included on all of my pages through PHP like so: <?php require_once “inc/footer.php”; ?> In the footer, I have a back to top button which I don’t want on some of my pages since they are small pages. What I have done so far is shown below. CSS (applying to all pages):
jQuery custom attr is not working in option tag
I have select type with having custom attr in option tag. When i am able to get the val, but not able to get the value in custom attribute. Here is Jsfiddle JSFIDDLE Answer So the problem is that you’re trying to get the values of the attributes from the select itself. What you want to do is to get
Calling class method in jquery
Hey im trying to build a simple inventory program. I have class like this : now i want to append rows in my table which contain select option like this: but new row wont added if call my class using normal php line. It does however add a new row if i remove the php line. So, how can i