sql面试必会6题经典 sql经典面试题及答案

如今姐姐们对于sql面试必会6题经典到底是怎么个情况？，姐姐们都需要分析一下sql面试必会6题经典，那么悠悠也在网络上收集了一些对于sql经典面试题及答案的一些内容来分享给姐姐们，原因是这样，姐姐们一起来看看吧。

恩,通过自定义函数吧.. 实现如下: ----创建自定义函数 create function F_Getvarchar(@a1 int) returns varchar(8000) as Begin declare @a2 varchar(100),@Newvarhar .

select *,a.过关数/a.总数 as 过关率 from( SELECT a.ID, COUNT(*) AS 总数,SUM(CASE b.STYLE WHEN 0 THEN 1 ELSE 0 E.

1、忍不住想说一句,因为第一题中的字段类型是 【日期型】,而各种数据库操作日. 没有一种共通的方法,所以脱离了数据库而言没有一种共通的sql. 2、select ID,.

select 政党id ,政党名称 ,党员人数 ,count(议员id ) from a,b where a.政党id = b.政党id order by count(议员id) desc

1.恐怕没有那么复杂,除了实体明细不论,只需要一个表 X(路线、站台)SQL语句是 select 路线 from x where 站台=站台1 and exists (select * from x x1 where x1.路线=路.

A.创建表格CODE省略 注明:学生表PK stu_id 课程表pk cos_id 分数表PK enrollment_id FK stu_id,cos_id B.插入数据code省略 C.Query 1. select s.stu_id,stu_name,count(cos_id) from student s,enrollments e where s.stu_id = e.stu_id and e.grade>60 group by s.stu_id,stu_name; 2. select e.stu_id,s.stu_name,c.cos_name from student s,enrollments e,course c where s.stu_id = e.stu_id and e.cos_id = c.cos_id and c.cos_name = 'CHINESE' and s.stu_name .

第一题,最高的10-20. 1234 select * from (select colD,colE,dense_rank() over(order by colG desc) as rk from tableb) where rk between 10 and 20 --当然面试的话这里可以注明 dense_rank 和 rank区别等等. 第二题,2楼写的就可以了 第三题,注意题目是人数之和 12345 select count(*) as cnt from tableB where colA in (select colA from tableA start with colA = 传入idconnect by prior colA = colB) 第四题 --测试数据 with tableA as (select 1 as colA,0 as .

1,select e_name from employee where e_hiredate>to_date('2001-01-01','YYYY-MM-dd') and e_hiredate<to_date('2003-01-01','YYYY-MM-dd') and dept_id in(select dept_id from depar where comp_id in (select comp_id from comp where comp_name='baidu'));<br>2,update employee set e_level=e_level+1,e_hiredate=sysdate where dept_id in(select dept_id from deptar where dept_name='监察部')

5.1 select a.username,b.deptname from users a,dept b where a.dept_id=b.id; 5.2 update users set dept_id='9' where dept_id='2'; 5.3 select a.deptname,b.count_id from dept a,(select dept_id,count(id) as count_id from users group by dept_id having count(id)>1) b where a.id=b.dept_id; 5.4 select a.deptname,b.count_man,c.count_woman from dept a,(select dept_id,count(sex) as count_man from users where sex='男' group by dept_id) b,(select dept_id,count(.

游标 存储过程 自定义函数 递归 ....

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