SQL进阶系列之10HAVING子句又回来了

写在前面

HAVING子句的处理对象是集合而不是记录

各队，全队点名

--各队，全体点名！
CREATE TABLE Teams
(member  CHAR(12) NOT NULL PRIMARY KEY,
 team_id INTEGER  NOT NULL,
 status  CHAR(8)  NOT NULL);

INSERT INTO Teams VALUES('乔',     1, '待命');
INSERT INTO Teams VALUES('肯',     1, '出勤中');
INSERT INTO Teams VALUES('米克',   1, '待命');
INSERT INTO Teams VALUES('卡伦',   2, '出勤中');
INSERT INTO Teams VALUES('凯斯',   2, '休息');
INSERT INTO Teams VALUES('简',     3, '待命');
INSERT INTO Teams VALUES('哈特',   3, '待命');
INSERT INTO Teams VALUES('迪克',   3, '待命');
INSERT INTO Teams VALUES('贝斯',   4, '待命');
INSERT INTO Teams VALUES('阿伦',   5, '出勤中');
INSERT INTO Teams VALUES('罗伯特', 5, '休息');
INSERT INTO Teams VALUES('卡根',   5, '待命');

-- 找出全队全员处于待命状态的队伍
-- NOT EXISTS法
SELECT team_id,member FROM Teams AS T1
WHERE NOT EXISTS (SELECT * FROM Teams AS T2 WHERE T1.team_id = T2.team_id AND status <> '待命' );
-- HAVING方法
SELECT team_id FROM Teams GROUP BY team_id HAVING COUNT(*) = SUM(CASE WHEN status = '待命' THEN 1 ELSE 0 END);
-- 变通方法1
SELECT team_id FROM Teams GROUP BY team_id HAVING MAX(status) = '待命' AND MIN(status) ='待命';
-- 变通方法2
SELECT team_id, CASE WHEN MAX(status) = '待命' AND MIN(status) = '待命' THEN '全都在待命' ELSE '队长！人手不够' END AS status
FROM Teams
GROUP BY team_id;

单重集合与多重集合

--单重集合与多重集合
CREATE TABLE Materials
(center         CHAR(12) NOT NULL,
 receive_date   DATE     NOT NULL,
 material       CHAR(12) NOT NULL,
 PRIMARY KEY(center, receive_date));

INSERT INTO Materials VALUES('东京'	,'2007-4-01',	'锡');
INSERT INTO Materials VALUES('东京'	,'2007-4-12',	'锌');
INSERT INTO Materials VALUES('东京'	,'2007-5-17',	'铝');
INSERT INTO Materials VALUES('东京'	,'2007-5-20',	'锌');
INSERT INTO Materials VALUES('大阪'	,'2007-4-20',	'铜');
INSERT INTO Materials VALUES('大阪'	,'2007-4-22',	'镍');
INSERT INTO Materials VALUES('大阪'	,'2007-4-29',	'铅');
INSERT INTO Materials VALUES('名古屋',	'2007-3-15',	'钛');
INSERT INTO Materials VALUES('名古屋',	'2007-4-01',	'钢');
INSERT INTO Materials VALUES('名古屋',	'2007-4-24',	'钢');
INSERT INTO Materials VALUES('名古屋',	'2007-5-02',	'镁');
INSERT INTO Materials VALUES('名古屋',	'2007-5-10',	'钛');
INSERT INTO Materials VALUES('福冈'	,'2007-5-10',	'锌');
INSERT INTO Materials VALUES('福冈'	,'2007-5-28',	'锡');

-- 找出原料出现重复的产地
SELECT center
FROM Materials
GROUP BY center
HAVING COUNT(*) <> COUNT(DISTINCT material);

-- 另一种写法
SELECT center,CASE WHEN COUNT(*) <> COUNT(DISTINCT material) THEN '存在重复' ELSE '不存在重复' END AS status
FROM Materials
GROUP BY center;

-- EXISTS写法
SELECT center,material
FROM Materials AS M1
WHERE EXISTS (SELECT * FROM Materials AS M2 WHERE M1.center = M2.center AND M1.receive_date <> M2.receive_date AND M1.material = M2.material);

寻找缺失的编号：升级版

--寻找缺失的编号：升级版
CREATE TABLE SeqTbl
( seq INTEGER NOT NULL PRIMARY KEY);

--不存在缺失编号（起始值＝1）
DELETE FROM SeqTbl;
INSERT INTO SeqTbl VALUES(1);
INSERT INTO SeqTbl VALUES(2);
INSERT INTO SeqTbl VALUES(3);
INSERT INTO SeqTbl VALUES(4);
INSERT INTO SeqTbl VALUES(5);

-- 如果有查询结果，说明存在缺失的编号
SELECT '存在缺失的编号' AS gap FROM SeqTbl HAVING COUNT(*) <> MAX(seq) --只适用于从1开始的序列
-- 如果有查询结果，说明存在缺失的编号：只调查数列的连续性
SELECT '存在缺失的编号' AS gap FROM SeqTbl HAVING COUNT(*) <> MAX(seq) - MIN(seq) + 1;
-- 不论是否存在都返回一行结果
SELECT CASE WHEN COUNT(*) = 0 THEN '空表'
            WHEN COUNT(*) = MAX(seq) - MIN(seq) + 1 THEN '不存在缺失编号' 
            ELSE '存在缺失编号' END AS gap
FROM SeqTbl;

-- 查找最小的缺失编号
SELECT CASE WHEN COUNT(*) = 0 OR MIN(seq) > 1 THEN 1
            ELSE (SELECT MIN(seq+1) FROM SeqTbl AS S1 WHERE NOT EXISTS (SELECT * FROM SeqTbl AS S2 WHERE S2.seq = S1.seq + 1)) END
FROM SeqTbl;

为集合设置详细的条件

--为集合设置详细的条件
CREATE TABLE TestResults
(student CHAR(12) NOT NULL PRIMARY KEY,
 class   CHAR(1)  NOT NULL,
 sex     CHAR(1)  NOT NULL,
 score   INTEGER  NOT NULL);

INSERT INTO TestResults VALUES('001', 'A', '男', 100);
INSERT INTO TestResults VALUES('002', 'A', '女', 100);
INSERT INTO TestResults VALUES('003', 'A', '女',  49);
INSERT INTO TestResults VALUES('004', 'A', '男',  30);
INSERT INTO TestResults VALUES('005', 'B', '女', 100);
INSERT INTO TestResults VALUES('006', 'B', '男',  92);
INSERT INTO TestResults VALUES('007', 'B', '男',  80);
INSERT INTO TestResults VALUES('008', 'B', '男',  80);
INSERT INTO TestResults VALUES('009', 'B', '女',  10);
INSERT INTO TestResults VALUES('010', 'C', '男',  92);
INSERT INTO TestResults VALUES('011', 'C', '男',  80);
INSERT INTO TestResults VALUES('012', 'C', '女',  21);
INSERT INTO TestResults VALUES('013', 'D', '女', 100);
INSERT INTO TestResults VALUES('014', 'D', '女',   0);
INSERT INTO TestResults VALUES('015', 'D', '女',   0);

-- 查出75%以上的学生分数都在80分以上的班级
SELECT class
FROM TestResults
GROUP BY class
HAVING COUNT(*) * 0.75 <= SUM(CASE WHEN score >= 80 THEN 1 ELSE 0 END);

-- 查出分数在50分以上的男生比女生多的班级
SELECT class
FROM TestResults
GROUP BY class
HAVING SUM(CASE WHEN score >= 50 AND sex = '男' THEN 1 ELSE 0 END) > SUM(CASE WHEN score >= 50 AND sex = '女' THEN 1 ELSE 0 END);

-- 比较男生和女生平均分的SQL语句(2):对空集求平均值后返回NULL
SELECT class
FROM TestResults
GROUP BY class
HAVING AVG(CASE WHEN sex = '男' THEN score ELSE NULL END) < AVG(CASE WHEN sex = '女' THEN score ELSE NULL END);

小结

常用调查集合性质的条件极其用途

No	条件表达式	用途
1	COUNT(DISTINCT col)=COUNT(col)	col列没有重复值
2	COUNT(*)=COUNT(col)	col列没有空值
3	COUNT(*)=MAX(col)	col列没有缺失(起始是1)
4	COUNT(*)=MAX(col)-MIN(col)+1	col列是连续的编号
5	MIN(col)=MAX(col)	col列是相同值或NULL
6	MIN(col)*MAX(col)>0	col列全是正数或全是负数
7	MIN(col)*MAX(col)<0	col列的最大值是正数，最小值是负数
8	MIN(ABS(col))	col最少有一个0
9	MIN(col-常量) = -MAX(col-常量)	col列的最大值和最小值与指定常量等距

• SQL指定搜索条件时，最重要的是搞清楚搜索的实体是集合还是集合的元素
• 如果一个实体对应着一行数据，那么就是元素，所以使用WHERE子句
• 如果一个实体对应着多行数据，那么就是集合，所以使用HAVING子句
• HAVING子句 可以通过聚合函数针对集合指定各种条件
• 如果通过CASE表达式生成特征函数，那么无论多么复杂的条件都可以描述
• HAVING子句很强大

练习题

/* 练习题1-10-1:单重集合与多重集合的一般化 */
CREATE TABLE Materials2
(center VARCHAR(32) NOT NULL,
 receive_date DATE  NOT NULL,
 material VARCHAR(32) NOT NULL,
 orgland  VARCHAR(32) NOT NULL,
 PRIMARY KEY(center, receive_date, material));

INSERT INTO Materials2 VALUES('东京',   '2007-04-01', '锡', 	'智利');
INSERT INTO Materials2 VALUES('东京',   '2007-04-12', '锌', 	'泰国');
INSERT INTO Materials2 VALUES('东京',   '2007-05-17', '铝', 	'巴西');
INSERT INTO Materials2 VALUES('东京',   '2007-05-20', '锌', 	'泰国');
INSERT INTO Materials2 VALUES('大阪',   '2007-04-20', '铜', 	'澳大利亚');
INSERT INTO Materials2 VALUES('大阪',   '2007-04-22', '镍', 	'南非');
INSERT INTO Materials2 VALUES('大阪',   '2007-04-29', '铅', 	'印度');
INSERT INTO Materials2 VALUES('名古屋', '2007-03-15', '钛', 	'玻利维亚');
INSERT INTO Materials2 VALUES('名古屋', '2007-04-01', '钢', 	'智利');
INSERT INTO Materials2 VALUES('名古屋', '2007-04-24', '钢', 	'阿根廷');
INSERT INTO Materials2 VALUES('名古屋', '2007-05-02', '镁', 	'智利');
INSERT INTO Materials2 VALUES('名古屋', '2007-05-10', '钛', 	'泰国');
INSERT INTO Materials2 VALUES('福冈',   '2007-05-10', '锌', 	'美国');
INSERT INTO Materials2 VALUES('福冈',   '2007-05-28', '锡',	'俄罗斯');

/* 练习题1-10-1：单重集合与多重集合的一般化 
-- 找出产品和原产地都一样的center
SELECT center
FROM Materials2 AS M1
WHERE EXISTS (SELECT * FROM Materials2 AS M2 WHERE M1.center = M2.center AND M1.receive_date <> M2.receive_date AND M1.material = M2.material AND M1.orgland = M2.orgland )
GROUP BY center;

/* 练习题1-10-1：单重集合与多重集合的一般化 
 选择（材料, 原产国）组合有重复的生产地 */
SELECT center
  FROM Materials2
 GROUP BY center
HAVING COUNT(material || orgland) <> COUNT(DISTINCT material || orgland);

/* 练习题1-10-2：多个条件的特征函数 */
CREATE TABLE TestScores
 (student_id INTEGER NOT NULL,
  subject    VARCHAR(16) NOT NULL,
  score      INTEGER NOT NULL,
    PRIMARY KEY (student_id, subject));

INSERT INTO TestScores VALUES(100, '数学', 100);
INSERT INTO TestScores VALUES(100, '语文', 80);
INSERT INTO TestScores VALUES(100, '理化', 80);
INSERT INTO TestScores VALUES(200, '数学', 80);
INSERT INTO TestScores VALUES(200, '语文', 95);
INSERT INTO TestScores VALUES(300, '数学', 40);
INSERT INTO TestScores VALUES(300, '语文', 50);
INSERT INTO TestScores VALUES(300, '社会', 55);
INSERT INTO TestScores VALUES(400, '数学', 80);

-- 找出数学分数在80分以上，且语文分数在50分以上的学生
SELECT student_id
FROM TestScores
WHERE subject IN ('语文','数学')
GROUP BY student_id
HAVING SUM(CASE WHEN subject = '数学' AND score > 80 THEN 1
                WHEN subject = '语文' AND score > 50 THEN 1
                ELSE 0 END)=2;