[Solved] Mysql: How do I subtract a percentage from a number
Try this: SELECT amount – (amount * discount_percentage / 100) FROM tableA; 0 solved Mysql: How do I subtract a percentage from a number
Try this: SELECT amount – (amount * discount_percentage / 100) FROM tableA; 0 solved Mysql: How do I subtract a percentage from a number
Your SQL CTE syntax is incorrect: you have a comma after the last CTE’s close-paren. The only time you need a comma there is if you have another CTE following it. Try changing to with unit as ( select t.RefDate, … ), — yes comma here, between CTEs agg as ( select u.RefDate, … ) … Read more
First, let’s define what we need. We need a table (view) with these fields: ProductId, ProductName, ProductViewCount, and RootCategoryId. Imagine, the Category table has RootCategoryId field already. Then we can use this query to receive the result: SELECT P.Id AS ‘ProductId’, P.Name AS ‘ProductName’, PVC.ProductViewCount, C.RootCategoryId FROM Category C INNER JOIN ProductCategory PC ON PC.CategoryId … Read more
This is one of the simplest types of SQL queries. SELECT user_id FROM yourTable WHERE user_name=”UserA” The first line says which columns you want to return in the result. The second line identifies the table you want to get the columns from. The third line specifies conditions on the rows you want to use from … Read more
If you just want to return the rows from each table that match, then you can use a UNION ALL: SELECT * FROM Table1 WHERE column2 = 1 union all SELECT * FROM Table2 WHERE column22= 1; See SQL Fiddle with Demo If you want to return the rows that match both criteria, then possibly … Read more
Here is a function that works fine for all cases and return the list of all first candidates encountered if no choice can be made to separate them. def find_best(list_of_lists): i = 0 while len(list_of_lists[i]) == 0: i+=1 list_containing_candidates = list_of_lists[i][:] if len(list_containing_candidates) == 1 : return list_containing_candidates[0] else: if i+1 < len(list_of_lists): for next_list … Read more
When I need a list of integers I use SELECT help_keyword_id num FROM mysql.help_keyword HAVING num BETWEEN 1 AND 100 ORDER BY 1; solved SELECT mysql query to get first 100 numbers (1-100) [closed]
Is this what you need? select { border: 0 none; color: black; background: transparent; font-size: 14px; padding: 6px; width: 100%; background: #58B14C; text-indent: 50%; } #mainselection { overflow: hidden; width: 100%; background: #4CAF50; text-align: center; } select:hover { text-shadow: 1px 1px red; box-shadow:1px 1px red; } <div id=”mainselection”> <select> <option>Select options</option> <option>1</option> <option>2</option> </select> </div> … Read more
There’s two ways to do this: (1) a session, and (2) a form. A session is used to store values between page requests – it uses a cookie to identify your browser, which PHP manages for you. Your category could be set into a variable, and then re-read at page load. The other approach is … Read more
You can use the MAX() and MIN() function to get the ids having largest and the id having smallest discount. Select id,discount from customer where discount=(select MAX(discount) from customer) OR discount=(select MIN(discount) from customer); 3 solved select id by the min and max (sql) [closed]