Sql Query用于增加多个项目的项目价格

时间:2014-12-08 08:03:04

标签: mysql sql select sql-update case

我想编写Sql Query以按百分比增加商品价格。

情景是: -

在表格中,我有3个coloumn:ID,Item-Name,Price

Example : If item-Name is T-shirt, Increase price by 10%

         item-Name is Jins , Increase price by 50%

         item-Name is top , Increase price by 5%

3 个答案:

答案 0 :(得分:2)

如果您要更新表格,可以进行条件更新。

update table_name
set 
price = 
case 
 when `Item-Name` = 'T-shirt' then price+( (price*10) /100 )
 when `Item-Name` = 'Jins' then price+( (price*50) /100 )
 when `Item-Name` = 'top' then price+( (price*5) /100 )
end ;

如果您希望在选择时没有在表格中进行任何更新来显示增加的价格,那么您可以执行以下操作。

select id,`Item-Name`,price,
case 
     when `Item-Name` = 'T-shirt' then price+( (price*10) /100 )
     when `Item-Name` = 'Jins' then price+( (price*50) /100 )
     when `Item-Name` = 'top' then price+( (price*5) /100 )
     else price
    end as new_price from table_name;

答案 1 :(得分:1)

试试这个:

SELECT a.ID, a.ItemName, a.Price, 
        (CASE WHEN a.ItemName = 'T-shirt' THEN (a.price * 10 / 100) 
              WHEN a.ItemName = 'Jins' THEN (a.price * 50 / 100) 
              WHEN a.ItemName = 'top' THEN (a.price * 5 / 100) 
              ELSE a.price 
         END) AS calculatedPrice
FROM tableA a 

答案 2 :(得分:0)

UPDATE TABLENAME SET price = (price*1.1) where item-Name = "T-shirt";

UPDATE TABLENAME SET price = (price*1.5) where item-Name = "Jins";

UPDATE TABLENAME SET price = (price*1.05) where item-Name = "Top";

我认为它有效......