关于group by 的一个SQL题目

获得部门最高工资比别的部门平均工资还低的部门及平均公司信息。

表如下： employee table

+----+--------+--------+------+

| id | name | salary | dept |

+----+--------+--------+------+

| 1 | test1 | 99 | dep1 |

| 2 | test2 | 89 | dep1 |

| 3 | test3 | 73 | dep1 |

| 4 | test4 | 87 | dep1 |

| 5 | test5 | 85 | dep1 |

| 6 | test6 | 84 | dep2 |

| 7 | test7 | 82 | dep2 |

| 8 | test8 | 80 | dep2 |

| 9 | test9 | 86 | dep2 |

| 10 | test10 | 79 | dep2 |

| 11 | test11 | 83 | dep3 |

| 12 | test12 | 80 | dep3 |

+----+--------+--------+------+

12 rows in set (0.00 sec)

有两个sql解决方案：

1，select a.*,b.* from

(select dept,max(salary) maxsalary from test group by dept) a,

(select dept,avg(salary) avgsalary from test group by dept) b

where a.maxsalary<b.avgsalary

2，select dept, max(salary) from employee group by dept having max(salary) in (select max(salary) from employee group by dept having max(salary) < some (select avg(salary) from employee group by dept))

第二题， 写一个SQL语句用来检查某个字段中是否存在重复的值,并输出重复的值和次数。

select dept, count(dept) from test.employee group by dept

第三题 横表变竖表

第四题，oracle如何更新一个表的sequence

alter seq increment by 1000;

select seq.nextval from seq;

alter seq increment by 1;

第五题， 如果更新一个表中的时间字段

update employee set start_date = SYSDATE - 5 where id = 1234;

or update employee set start_date = to_date('2003-06-13 15:18','YYYY-MM-DD hh:mm');