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新题速看!2021阿里、腾讯、字节都在问的SQL数据库笔试题及答案都给你整理好啦!

2021-12-07 16:32:48  阅读:206  来源: 互联网

标签:题及 -% 新题 st so SQL BaseOrder where select


 ♥  前  言

2021到了最后一个月份,年后肯定有蛮多小伙伴需要跳槽换工作,但对于年限稍短的软件测试工程师,难免会需要进行笔试,而在笔试中,基本都会碰到一道关于数据库的大题,今天这篇文章呢,就收录了下最近学员反馈上来的一些数据库笔试真题,包含答案!

 

1、依据以下学生表、班级表,按要求写 SQL

 

图片

 

答案:

1、
select * from student a
left join class b on a.c_id = b.id
where score = (
select max(score) from student)

2、
select b.name,count(*),avg(score) as avgScore from student a,class b
where a.c_id = b.id
group by b.name
order by avgScore;

 

2、Table A 保存客户的基本信息;Table B 保存客户的资产负债信息。按要求写 SQL

 

图片

 

答案:

(1)select ID from A where name = '李四';
(2)select NO,NAME,ID from A,B where A.NO = B.NO AND DEPOSIT >= 1000;
(3)select COUNTRY,count(*) FROM A GROUP BY COUNTRY;

(4)SELECT '80 后' as 年龄段,sum(CREDIT) AS '信用卡余额' FROM A,B WHERE A.NO = B.NO AND BIRTH >= 19800101 AND BIRTH < 19900101
UNION
SELECT '90 后' as 年龄段,sum(CREDIT) AS '信用卡余额' FROM A,B WHERE A.NO = B.NO AND BIRTH >= 19900101 AND BIRTH < 20000101;

 

3、数据库(编写示例 sql)

 

图片

图片

 

答案:

  1. select orderNo, if(status=1,'新建','处理中') from OrderTrans;

  2. select DATE_FORMAT(a.crttime,'%m-%d-%Y') ,username,count(*) from testCase a,User b where a.crtUser = b.UserId group by DATE_FORMAT(a.crttime,'%m-%d-%Y'),b.username, having status = '成功';

  3. select count(*),sum(amount),DATE_FORMAT(crttime,'%m-%d-%Y') from OrderInfo group by
    DATE_FORMAT(crttime,'%m-%d-%Y');

 

4、现有三张数据表如下:学生资料表:记录学生基本信息;课程表:记录课程基本信息;成绩表:记录每人各门课程成绩的信息,1 个学生对应多个成绩,1 个成绩只属于一个学生,一个课程

 

图片

 

答案:

  1. Select * from students where jg = ‘湖北’ and birthday = ‘1992-6-1’ order by no asc;

  2. Select avg(so.cj),min(so.cj),max(so.cj),sum(so.cj) from student st,course c,source so where st.no = so.no and c.kebh=so.kebh and st.name = ‘王华’ group by st.no;

  3. Select st.no,st.name,st.bj,c.kebh,c.kcmc,so.cj from student st,course c,source so where st.no = so.no and c.kebh=so.kebh and st.name = ‘张三’ order by so.cj desc;

 

图片

图片

 

答案:

  1. Select Consultant.SubsidiaryID, BaseOrder.* from Consultant, BaseOrder where Consultant.ConsultantID = BaseOrder.ConsultantID and Consultant.SubsidiaryID = 29 order by Orderid desc;

  2. Select BaseOrder.ConsultantID,sum(OrderDetails.TotalPrice) from BaseOrder,OrderDetails where BaseOrder.Orderid=OrderDetails.Orderid and month(OrderDate) = 5 group by BaseOrder.ConsultantID;

  3. Insert into Consultant (ConsultantID,ConstultantStatusID,SubsidiaryID,Name) values (200000,10,29,’Gary’);

  4. Delete from Consultant where Name like ‘%Gary%’;

  5. Update BaseOrder set OrderDate = sysdate() where ConsultantID in (select ConsultantID from BaseOrder where ConsultantID=100003 order by OrderDate desc limit 0,1);

 

 

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标签:题及,-%,新题,st,so,SQL,BaseOrder,where,select
来源: https://www.cnblogs.com/tudou-22/p/15656794.html

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