写在前面
关系模型的数据结构里，并没有顺序的概念，但SQL处理有序集合也有坚实的理论基础
生成连续编号
--生成连续编号
CREATE TABLE Digits
(digit INTEGER PRIMARY KEY);
INSERT INTO Digits VALUES (0);
INSERT INTO Digits VALUES (1);
INSERT INTO Digits VALUES (2);
INSERT INTO Digits VALUES (3);
INSERT INTO Digits VALUES (4);
INSERT INTO Digits VALUES (5);
INSERT INTO Digits VALUES (6);
INSERT INTO Digits VALUES (7);
INSERT INTO Digits VALUES (8);
INSERT INTO Digits VALUES (9);-- 创建00-99的序列
SELECT (D1.digit + D2.digit * 10) AS seq
FROM Digits AS D1 CROSS JOIN Digits AS D2
ORDER BY seq;
-- 创建1到542的序列
SELECT (D1.digit + D2.digit * 10 + D3.digit * 100) AS seq
FROM Digits AS D1,Digits AS D2,Digits AS D3
WHERE (D1.digit + D2.digit * 10 + D3.digit * 100) BETWEEN 1 AND 542
ORDER BY seq;CREATE VIEW Sequence (seq) AS
SELECT D1.digit + (D2.digit * 10 ) + (D3.digit * 100)
FROM Digits D1,Digits D2,Digits D3;
SELECT seq FROM Sequence WHERE seq BETWEEN 1 AND 100 ORDER BY seq;求全部的缺失编号
-- EXCEPT版
SELECT seq FROM Sequence WHERE seq BETWEEN 1 AND 12
EXCEPT (SELECT * FROM SeqTbl);
-- NOT IN版
SELECT seq FROM Sequence WHERE seq BETWEEN 1 AND 12 AND seq
NOT IN (SELECT * FROM SeqTbl);-- 动态地指定连续编号范围的SQL语句
SELECT seq FROM Sequence WHERE seq BETWEEN (SELECT MIN(seq) FROM SeqTbl) AND (SELECT MAX(seq) FROM SeqTbl) EXCEPT (SELECT * FROM SeqTbl);三个人能坐得下吗？
--三个人能坐得下吗？
CREATE TABLE Seats
( seat INTEGER NOT NULL PRIMARY KEY,
status CHAR(6) NOT NULL
CHECK (status IN ('未预订', '已预订')) );
INSERT INTO Seats VALUES (1, '已预订');
INSERT INTO Seats VALUES (2, '已预订');
INSERT INTO Seats VALUES (3, '未预订');
INSERT INTO Seats VALUES (4, '未预订');
INSERT INTO Seats VALUES (5, '未预订');
INSERT INTO Seats VALUES (6, '已预订');
INSERT INTO Seats VALUES (7, '未预订');
INSERT INTO Seats VALUES (8, '未预订');
INSERT INTO Seats VALUES (9, '未预订');
INSERT INTO Seats VALUES (10, '未预订');
INSERT INTO Seats VALUES (11, '未预订');
INSERT INTO Seats VALUES (12, '已预订');
INSERT INTO Seats VALUES (13, '已预订');
INSERT INTO Seats VALUES (14, '未预订');
INSERT INTO Seats VALUES (15, '未预订');-- 找出需要的空位(1):不考虑座位的换排
SELECT S1.seat AS start_seat,'~',S2.seat AS end_seat
FROM Seats S1,Seats S2
WHERE S2.seat = S1.seat + 2 AND NOT EXISTS
(SELECT * FROM Seats S3 WHERE S3.seat BETWEEN S1.seat AND S2.seat AND S3.status <> '未预订');--考虑座位的折返
CREATE TABLE Seats2
( seat INTEGER NOT NULL PRIMARY KEY,
row_id CHAR(1) NOT NULL,
status CHAR(6) NOT NULL
CHECK (status IN ('未预订', '已预订')) );
INSERT INTO Seats2 VALUES (1, 'A', '已预订');
INSERT INTO Seats2 VALUES (2, 'A', '已预订');
INSERT INTO Seats2 VALUES (3, 'A', '未预订');
INSERT INTO Seats2 VALUES (4, 'A', '未预订');
INSERT INTO Seats2 VALUES (5, 'A', '未预订');
INSERT INTO Seats2 VALUES (6, 'B', '已预订');
INSERT INTO Seats2 VALUES (7, 'B', '已预订');
INSERT INTO Seats2 VALUES (8, 'B', '未预订');
INSERT INTO Seats2 VALUES (9, 'B', '未预订');
INSERT INTO Seats2 VALUES (10,'B', '未预订');
INSERT INTO Seats2 VALUES (11,'C', '未预订');
INSERT INTO Seats2 VALUES (12,'C', '未预订');
INSERT INTO Seats2 VALUES (13,'C', '未预订');
INSERT INTO Seats2 VALUES (14,'C', '已预订');
INSERT INTO Seats2 VALUES (15,'C', '未预订');-- 找出需要的空位(2):考虑座位的换排
SELECT S1.seat AS start_seat,'~',S2.seat AS end_seat
FROM Seats2 S1,Seats2 S2
WHERE S2.seat = S1.seat + 2 AND NOT EXISTS
(SELECT * FROM Seats2 S3 WHERE S3.seat BETWEEN S1.seat AND S2.seat AND (S3.status <> '未预订' OR S3.row_id <> S1.row_id));最多能坐下多少人
--最多能坐下多少人？
CREATE TABLE Seats3
( seat INTEGER NOT NULL PRIMARY KEY,
status CHAR(6) NOT NULL
CHECK (status IN ('未预订', '已预订')) );
INSERT INTO Seats3 VALUES (1, '已预订');
INSERT INTO Seats3 VALUES (2, '未预订');
INSERT INTO Seats3 VALUES (3, '未预订');
INSERT INTO Seats3 VALUES (4, '未预订');
INSERT INTO Seats3 VALUES (5, '未预订');
INSERT INTO Seats3 VALUES (6, '已预订');
INSERT INTO Seats3 VALUES (7, '未预订');
INSERT INTO Seats3 VALUES (8, '已预订');
INSERT INTO Seats3 VALUES (9, '未预订');
INSERT INTO Seats3 VALUES (10, '未预订');-- 先生成存储了所有序列的视图
CREATE VIEW Sequences(start_seat,end_seat,seat_cnt) AS
SELECT S1.seat AS start_seat,S2.seat AS end_seat,S2.seat-S1.seat + 1 AS seat_cnt
FROM Seats3 S1,Seats3 S2
WHERE S1.seat < S2.seat
AND NOT EXISTS (SELECT * FROM Seats3 S3 WHERE (S3.seat BETWEEN S1.seat AND S2.seat AND S3.status<> '未预订') OR (S3.seat = S2.seat + 1 AND S3.status = '未预订') OR (S3.seat = S1.seat - 1 AND S3.status = '未预订'));
-- 取出最长的序列
SELECT start_seat,end_seat,seat_cnt FROM
Sequences WHERE seat_cnt = (SELECT MAX(seat_cnt) FROM Sequences);单调递增和单调递减
--单调递增和单调递减
CREATE TABLE MyStock
(deal_date DATE PRIMARY KEY,
price INTEGER );
INSERT INTO MyStock VALUES ('2007-01-06', 1000);
INSERT INTO MyStock VALUES ('2007-01-08', 1050);
INSERT INTO MyStock VALUES ('2007-01-09', 1050);
INSERT INTO MyStock VALUES ('2007-01-12', 900);
INSERT INTO MyStock VALUES ('2007-01-13', 880);
INSERT INTO MyStock VALUES ('2007-01-14', 870);
INSERT INTO MyStock VALUES ('2007-01-16', 920);
INSERT INTO MyStock VALUES ('2007-01-17', 1000);-- 生成起点到终点的组合
SELECT My1.deal_date,My2.deal_date
FROM MyStock AS My1,MyStock AS My2
WHERE My1.deal_date < My2.deal_date
AND NOT EXISTS
(SELECT * FROM MyStock AS My3,MyStock AS My4
WHERE My3.deal_date BETWEEN My1.deal_date AND My2.deal_date
AND My3.deal_date BETWEEN My1.deal_date AND My2.deal_date
AND My3.deal_date < My4.deal_date
AND My3.price >= My4.price);--排除掉子集，只取最长的时间区间
SELECT MIN(start_date) AS start_date, /* 最大限度地向前延伸起点 */
end_date
FROM (SELECT S1.deal_date AS start_date,
MAX(S2.deal_date) AS end_date /* 最大限度地向后延伸终点 */
FROM MyStock S1, MyStock S2
WHERE S1.deal_date < S2.deal_date
AND NOT EXISTS
(SELECT *
FROM MyStock S3, MyStock S4
WHERE S3.deal_date BETWEEN S1.deal_date AND S2.deal_date
AND S4.deal_date BETWEEN S1.deal_date AND S2.deal_date
AND S3.deal_date < S4.deal_date
AND S3.price >= S4.price)
GROUP BY S1.deal_date) TMP
GROUP BY end_date
ORDER BY start_date;小结
- SQL处理数据的方法有两种
- 第一种把数据看成忽略了顺序的集合
- 第二种把数据看成有序的集合，此时的基本方法如下：
- 首先自连接生成起点到终点的集合
- 其次在子查询中描述内部的各个元素之间必须满足的关系
- 要在SQL中表达全称量化时，需要将全称量化命题转化为存在量化命题的否定形式，并使用NOT EXISTS谓词。
练习题
/* 练习题1-9-1：求所有的缺失编号——NOT EXISTS和外连接
NOT EXISTS版 */
SELECT seq
FROM Sequence N
WHERE seq BETWEEN 1 AND 12
AND NOT EXISTS
(SELECT *
FROM SeqTbl S
WHERE N.seq = S.seq );/* 练习题1-9-1：求所有的缺失编号——NOT EXISTS和外连接
NOT EXISTS版 */
SELECT N.seq
FROM Sequence N LEFT OUTER JOIN SeqTbl S
ON N.seq = S.seq
WHERE N.seq BETWEEN 1 AND 12
AND S.seq IS NULL;/* 练习题1-9-2：求序列——面向集合的思想 */
SELECT S1.seat AS start_seat, '～' , S2.seat AS end_seat
FROM Seats S1, Seats S2, Seats S3
WHERE S2.seat = S1.seat + (:head_cnt -1)
AND S3.seat BETWEEN S1.seat AND S2.seat
GROUP BY S1.seat, S2.seat
HAVING COUNT(*) = SUM(CASE WHEN S3.status = '未预订' THEN 1 ELSE 0 END);/* 坐位有换排时 */
SELECT S1.seat AS start_seat, ' ～ ' , S2.seat AS end_seat
FROM Seats2 S1, Seats2 S2, Seats2 S3
WHERE S2.seat = S1.seat + (:head_cnt -1)
AND S3.seat BETWEEN S1.seat AND S2.seat
GROUP BY S1.seat, S2.seat
HAVING COUNT(*) = SUM(CASE WHEN S3.status = '未预订'
AND S3.row_id = S1.row_id THEN 1 ELSE 0 END);
/* 练习题1-9-3：求所有的序列——面向集合的思想 */
SELECT S1.seat AS start_seat,
S2.seat AS end_seat,
S2.seat - S1.seat + 1 AS seat_cnt
FROM Seats3 S1, Seats3 S2, Seats3 S3
WHERE S1.seat <= S2.seat /* 第一步：生成起点和终点的组合 */
AND S3.seat BETWEEN S1.seat - 1 AND S2.seat + 1
GROUP BY S1.seat, S2.seat
HAVING COUNT(*) = SUM(CASE WHEN S3.seat BETWEEN S1.seat AND S2.seat
AND S3.status = '未预订' THEN 1 /* 条件1 */
WHEN S3.seat = S2.seat + 1 AND S3.status = '已预订' THEN 1 /* 条件2 */
WHEN S3.seat = S1.seat - 1 AND S3.status = '已预订' THEN 1 /* 条件3 */
ELSE 0 END);原文地址：https://www.cnblogs.com/evian-jeff/p/11605207.html
时间： 2024-11-10 05:12:16