不忽略空值null的聚合函数_sqlzoo 练习题答案聚合函数和空值部分

1 SUM and COUNT

1 显示世界总人口

SELECT SUM(population)
FROM world

2 列出所有的洲份, 每个只有一次。

SELECT DISTINCT continent
FROM world

3 找出非洲(Africa)的GDP总和。

SELECT SUM(gdp)
FROM world
WHERE continent = 'Africa'

4 有多少个国家具有至少百万(1000000)的面积。

SELECT COUNT(*)
FROM world
WHERE area >= 1000000

5 (‘France’,’Germany’,’Spain’)（“法国”，“德国”，“西班牙”）的总人口是多少？

SELECT SUM(population)
FROM world
WHERE name IN ('France','Germany','Spain')

6 对于每一个洲份，显示洲份和国家的数量。

SELECT continent,COUNT(*)
FROM world
GROUP BY continent

7 对于每一个洲份，显示洲份和至少有1000万人(10,000,000)口国家的数目。

SELECT continent,COUNT(*)
FROM world
WHERE population > 10000000
GROUP BY continent

8 列出有至少100百万(1亿)(100,000,000)人口的洲份。

SELECT continent
FROM world
GROUP BY continent
HAVING SUM(population) > 100000000

2 The nobel table can be used to practice more SUM and COUNT functions.

1 找出总共有多少个奖颁发了。

SELECT COUNT(*)
FROM nobel

2 列出每一个奖项(subject), 只列一次

SELECT DISTINCT subject
FROM nobel

3 找出物理奖的总颁发次数。

SELECT COUNT(subject)
FROM nobel
WHERE subject = 'Physics'

4 对每一个奖项(Subject),列出颁发数目。

SELECT subject,COUNT(*)
FROM nobel
GROUP BY subject

5 对每一个奖项(Subject),列出首次颁发的年份。

SELECT subject,yr
FROM nobel n1
WHERE yr <= (
SELECT MIN(yr) FROM nobel n2
WHERE n1.subject = n2.subject)
GROUP BY subject,yr

6 对每一个奖项(Subject),列出2000年颁发的数目。

SELECT subject,COUNT(*)
FROM nobel
WHERE yr = 2000
GROUP BY subject

7 对每一个奖项(Subject),列出有多少个不同的得奖者

SELECT subject,COUNT(DISTINCT winner)
FROM nobel
GROUP BY subject

这个题目关键在于列出不同的获奖者数量，由于可能存在一个人多次获得同一奖项，所以需要使用DISTINCT 去重。

8 对每一个奖项(Subject),列出有多少年曾颁发过。

SELECT subject,COUNT(DISTINCT yr)
FROM nobel
GROUP BY subject

9 列出哪年曾同年有3个物理奖Physics得奖者。

SELECT yr
FROM nobel
WHERE subject = 'Physics'
GROUP BY yr
HAVING COUNT(winner)=3

10 列出谁得奖多于一次。

SELECT winner
FROM nobel
GROUP BY winner
HAVING COUNT(*) > 1

11 列出谁获得多于一个奖项(Subject)

SELECT winner
FROM nobel
GROUP BY winner
HAVING COUNT(DISTINCT subject) > 1

12 哪年哪奖项，是同一奖项(subject)颁发给3个人。只列出2000年及之后的资料。

SELECT yr,subject
FROM nobel
WHERE yr >=2000
GROUP BY yr,subject
HAVING COUNT(winner)=3

3 Using Null

1 列出学系department是NULL值的老师。

SELECT name
FROM teacher
WHERE dept IS NULL

2 注意INNER JOIN 不理会没有学系的老师及没有老师的学系。

SELECT teacher.name, dept.nameFROM teacher INNER JOIN deptON (teacher.dept=dept.id)

3 使用不同的JOIN(外连接)，来列出全部老师。

SELECT t.name,d.name
FROM teacher t
LEFT JOIN dept d ON t.dept = d.id

LEFT JOIN 也可以写成 LEFT OUTTER JOIN 两者相同

4 使用不同的JOIN(外连接)，来列出全部学系。

SELECT t.name,d.name
FROM teacher t
RIGHT JOIN dept d ON t.dept = d.id

5 Use COALESCE to print the mobile number. Use the number '07986 444 2266' if there is no number given.Show teacher name and mobile number or '07986 444 2266'

SELECT name, COALESCE(mobile, '07986 444 2266')
FROM teacher

6 Use the COALESCE function and a LEFT JOIN to print the teachernameand department name. Use the string 'None' where there is no department.

SELECT teacher.name, COALESCE(dept.name, 'None')
FROM teacher LEFT JOIN dept ON teacher.dept = dept.id

7 使用COUNT来计算老师和流动电话数目。

SELECT COUNT(name),COUNT(mobile)
FROM teacher

8 使用COUNT 和 GROUP BY dept.name来显示每一学系的老师数目。使用 RIGHT JOIN 以确保工程系Engineering 是在当中。

SELECT COUNT(name)
FROM teacher t
RIGHT JOIN dept d ON t.dept = d.id
GROUP BY dept,name

得出的结果没错只是和答案顺序不一样

9 Use CASE to show the name of each teacher followed by 'Sci' if the teacher is in dept 1 or 2 and 'Art' otherwise.

SELECT name,
CASE WHEN dept = 1 OR dept = 2 THEN 'Sci' ELSE 'Art' END
FROM teacher

10 Use CASE to show the name of each teacher followed by 'Sci' if the teacher is in dept 1 or 2, show 'Art' if the teacher's dept is 3 and 'None' otherwise.

SELECT name,
CASE WHEN dept = 1 OR dept = 2 THEN 'Sci'WHEN dept = 3 THEN 'Art'ELSE 'None' END
FROM teacher;

4 Scottish Parliament

1 一个成员被工党逐出党，现没属任何党。找出他

SELECT name
FROM msp
WHERE party IS NULL

2 列出每个党及其领导人。

SELECT name,leader
FROM party

3 列出每个党及其领导人，有些党其实是没有领导人的。

SELECT name,leader
FROM party
WHERE leader IS NOT NULL

4 列出政党名单，当中最少有一名党员在议会内。

SELECT name
FROM party
WHERE code IN (
SELECT party FROM msp )

或者

SELECT p.name
FROM msp m
JOIN party p ON p.code = m.party
WHERE m.name IS NOT NULL
GROUP BY p.name

错误我最开始使用如下代码

SELECT name
FROM party
WHERE leader IN (SELECT name FROM msp )

错误原因：party 中只显示了领导人，有可能该当在议会的成员为非领导人，所以结果不完整。

5 列出议会成员的名单，如有所属政党，一同列出。确保 Canavan MSP, Dennis 是在名单中。按msp.name顺序排列。

SELECT m.name,p.name
FROM msp m
LEFT JOIN party p ON p.code = m.party
ORDER BY m.name

6 列出议会中每一政党的党员人数。

SELECT p.Name,COUNT(*)
FROM msp m
JOIN party p on m.party = p.code
WHERE m.name IS NOT NULL
GROUP BY p.Name

7 列出每一政党的议会中党员人数，包括没有党员在议会中的政党。

SELECT p.name,COUNT(m.party)
FROM msp m
RIGHT JOIN party p ON p.code = m.party
GROUP BY p.name