SQL面试常见问题解析(上)

SQL面试常见问题解析(上)

1.用一条sql语句查出所有的课程得分大于80分的同学姓名。

name	subject	score
张三	语文	81
张三	数学	75
李四	语文	76
李四	数学	90
王五	语文	81
王五	数学	100
王五	英语	90

这道题目有两种解法：

从正面来看，可以以NAME分组然后每组的最低得分如果都大于80分，那么该同学就满足要求

SELECT NAME FROM students GROUP BY NAME HAVING MIN(score) > 80;

从反面来看，我们可以先选出所有得分低于80分的同学，然后选出所有不属于前面结果集的同学即可

SELECT DISTINCT(NAME) FROM students where name not in (SELECT NAME FROM students where score < 80);

2.删除除了自动编号不同, 其他都相同的学生冗余信息

auto_no	no	name	subject_no	subject	score
1	2018001	张三	0001	数学	69
2	2018002	李四	0002	语文	72
3	2018001	张三	0001	数学	69

这道题目就是删除不在分组(no, name, subject_no, subject, score)的行

DELETE FROM `school` where auto_no not in (SELECT MAX(auto_no) FROM `school` GROUP BY no, name, subject_no, subject, score);

DELETE FROM `school` where auto_no not in (SELECT MAX(t.auto_no) FROM (SELECT auto_no, no, name, subject_no, subject, score FROM `school` ) as t GROUP BY no, name, subject_no, subject, score);

3.一个叫team的表，里面只有一个字段name，一共有4条记录，分别是a,b,c,d对应四个球队，现在四个球队进行比赛，用一条sql语句显示所有的可能比赛组合。

表结构如下:

Name
a
b
c
d

执行下面代码就可以得到结果：

SELECT a.name, b.name FROM `team` AS a, `team` AS b where a.name != b.name;

name	name
b	a
c	a
d	a
a	b
c	b
d	b
a	c
b	c
d	c
a	d
b	d
c	d

4.请用SQL 语句实现：从JcyAudit 数据表中查询出所有月份的发生额都比101 科目相应月份的发生额高的科目。请注意：JcyAudit 中有很多科目，都有1 －12 月份的发生额。

AccID ：科目代码，Occmonth ：发生额月份，DebitOccur ：发生额。
数据库名：JcyAudit ，数据集：Select * from TestDB

解决sql语句复杂查询最好是要分步进行

1. 先查询出所有的AccID = 101的行

   SELECT * FROM JcyAudit WHERE AccID = 101;

   ​

2. 然后查询出所有相同月份但是发生额大于101科目的行

   SELECT a.Occmonth, a.AccID FROM JcyAudit AS a, (SELECT * FROM JcyAudit WHERE AccID = 101) AS b WHERE a.Occmonth = b.Occmonth AND a.DebitOccur > b.DebitOccur;

3. 最后统计上面的查询结果

   SELECT c.AccID FROM (
      SELECT a.Occmonth, a.AccID FROM JcyAudit AS a, (SELECT * FROM JcyAudit WHERE AccID = 101) AS b WHERE a.Occmonth = b.Occmonth AND a.DebitOccur > b.DebitOccur
   ) AS c GROUP BY c.AccID HAVING count(c.AccID) = 12;

   ​

5.面试题：怎么把这样一个表

year	month	amount
1991	1	1.1
1991	2	1.2
1991	3	1.3
1991	4	1.4
1992	1	2.1
1992	2	2.2
1992	3	2.3
1992	4	2.4

查成这样一个结果

year	m1	m2	m3	m4
1991	1.1	1.2	1.3	1.4
1992	2.1	2.2	2.3	2.4

SELECT `year`, 
(SELECT amount FROM total AS t WHERE t.year = total.year AND month = 1) AS m1,
(SELECT amount FROM total AS t WHERE t.year = total.year AND month = 2) AS m2,
(SELECT amount FROM total AS t WHERE t.year = total.year AND month = 3) AS m3,
(SELECT amount FROM total AS t WHERE t.year = total.year AND month = 4) AS m4
FROM total GROUP BY year;