您可以为此使用CASE语句并设置条件以在临时列中获取结果。
让我们首先创建一个表-
mysql> create table DemoTable ( EmployeeId int NOT NULL AUTO_INCREMENT PRIMARY KEY, EmployeeName varchar(20), EmployeeSalary int, EmployeeExperience int );
以下是使用insert命令在表中插入一些记录的查询-
mysql> insert into DemoTable(EmployeeName,EmployeeSalary,EmployeeExperience) values('Larry',4500,5); mysql> insert into DemoTable(EmployeeName,EmployeeSalary,EmployeeExperience) values('Mike',130000,8); mysql> insert into DemoTable(EmployeeName,EmployeeSalary,EmployeeExperience) values('Sam',11000,5); mysql> insert into DemoTable(EmployeeName,EmployeeSalary,EmployeeExperience) values('Carol',140000,8) ;
以下是查询以使用select命令显示表中的记录-
mysql> select *from DemoTable;
这将产生以下输出-
+------------+--------------+----------------+--------------------+ | EmployeeId | EmployeeName | EmployeeSalary | EmployeeExperience | +------------+--------------+----------------+--------------------+ | 1 | Larry | 4500 | 5 | | 2 | Mike | 130000 | 8 | | 3 | Sam | 11000 | 5 | | 4 | Carol | 140000 | 8 | +------------+--------------+----------------+--------------------+ 4 rows in set (0.00 sec)
以下是在MySQL中添加一个临时列的查询,其中值取决于另一列。这里的临时列是NewSalary-
mysql> select EmployeeId,EmployeeName,EmployeeSalary,EmployeeExperience, case when EmployeeExperience=5 then EmployeeSalary+10000 when EmployeeExperience=8 then EmployeeSalary+20000 else null end as NewSalary from DemoTable;
这将产生以下输出-
+------------+--------------+----------------+--------------------+-----------+ | EmployeeId | EmployeeName | EmployeeSalary | EmployeeExperience | NewSalary | +------------+--------------+----------------+--------------------+-----------+ | 1 | Larry | 4500 | 5 | 14500 | | 2 | Mike | 130000 | 8 | 150000 | | 3 | Sam | 11000 | 5 | 21000 | | 4 | Carol | 140000 | 8 | 160000 | +------------+--------------+----------------+--------------------+-----------+ 4 rows in set (0.00 sec)