I'm learning SQL and I try some things in phpMyAdmin like creating or deleting tables, creating or deleting records and so on... Let's assume I have a table where there are 2 columns - id and name populated with 7…
I'm going to delete some rows in one of my table called sy_user. I want to delete some of the rows that is not matched in the other table, which is the tb_master_pegawai. How do i solve this query to…
the thing is, i have a table students table and another scores table and i already have all the students in the scores table but the class field is still empty so, i thought i could update all of them…
Okay, so I'm working on this Microsoft SQL database table that contains message strings, and I am trying to identify all of the URLs in those strings via starting with 'www.' and ending with either ('.biz','.com','.net','.org'). I found a very…
In my prestashop database I tried to update the default product attribute when there is 0 quantity for that one. I check if quantity is 0 and then set default_on to null. Then I try to set another attribute default_on…
How can I perform a SQL query for the following? $args = array( 'customer_id' => $customer_id, 'status' => 'completed', 'limit' => 1, ); $orderslast = wc_get_orders( $args ); Here's an error I get: 2019/08/14 09:08:03 [error] 11#11: *19 FastCGI sent…
i have 3 tables like this questions_table question_id | content | user | 1 | my first question | [email protected] | 2 | my second question | [email protected]| replies_table reply_id | question_id|user |content |voteCount| 1 | 1 |[email protected] |first reply…
Thank to your help I made a view in my database called 'people' that retrieve data using three functions called 'isUserVerified', 'hasUserPicture' and 'userHobbies' from two tables called 'users' and 'user_hobbies': SELECT `u`.`id` AS `id`, `isUserVerified`(`u`.`id`) AS `verification`, `hasUserPicture`(`u`.id) AS…
I've set a local server using PHPMyAdmin, and I'm presenting some dynamic data that is stored in that server using some PHP, HTML and SQL. The problem is that whenever I pass a variable that is stored using $variable =…
When i try to create view like this CREATE VIEW data2tables AS SELECT * FROM Employees e INNER JOIN Computers c ON e.id = c.id WHERE e.name = 'Georgi' AND c.department = 'Sales' it gives me error saying that #1060…