数据库迭代查询作业(绝对能运行)
数据库迭代查询作业 emp(eno, ename, salary, mgr),其中mgr是员工领导,设计约束:要求领导的工资不能低于他的下属平均工资。(注意,这是一个递归查询) 下面是一些样例数据 CREATE TABLE dbo.Employees ( empid INT NOT NULL PRIMARY KEY, mgrid INT NULL REFERENCES dbo.Employees, empname VARCHAR(25) NOT NULL, salary MONEY NOT NULL, CHECK (empid <> mgrid) ); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(1, NULL, 'David', $10000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(2, 1, 'Eitan', $7000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(3, 1, 'Ina', $7500.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(4, 2, 'Seraph', $5000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(5, 2, 'Jiru', $5500.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(6, 2, 'Steve', $4500.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(7, 3, 'Aaron', $5000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(8, 5, 'Lilach', $3500.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(9, 7, 'Rita', $3000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(10, 5, 'Sean', $3000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(11, 7, 'Gabriel', $3000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(12, 9, 'Emilia' , $2000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(13, 9, 'Michael', $2000.00); INSERT INTO dbo.Employees(empid, mgrid, empname, salary) VALUES(14, 9, 'Didi', $1500.00);
- 码兴邦2014-05-27不错,很有借鉴价值
- yingzi12342014-05-30很不错的资源,运行成功了
- ouyuyao2015-08-10看了一下 能用
- phla_han2014-05-10这个必须顶,作业解决了~
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