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SQL进阶系列之8EXISTS谓词的用法

时间:2021-07-01 10:21:17 帮助过:65人阅读

都满足条件\(P\)
  • 存在量词:存在(至少有一个)满足条件\(P\)\(x\)
  • EXISTS谓词实现了存在量词(因此,可以根据德摩根律实现全称量化)
  • 实践篇

    查询表中不存在的数据

    /* 查询表中“不”存在的数据 */
    CREATE TABLE Meetings
    (meeting CHAR(32) NOT NULL,
     person  CHAR(32) NOT NULL,
     PRIMARY KEY (meeting, person));
    
    INSERT INTO Meetings VALUES('第1次', '伊藤');
    INSERT INTO Meetings VALUES('第1次', '水岛');
    INSERT INTO Meetings VALUES('第1次', '坂东');
    INSERT INTO Meetings VALUES('第2次', '伊藤');
    INSERT INTO Meetings VALUES('第2次', '宫田');
    INSERT INTO Meetings VALUES('第3次', '坂东');
    INSERT INTO Meetings VALUES('第3次', '水岛');
    INSERT INTO Meetings VALUES('第3次', '宫田');
    -- 求所有人参加所有会的笛卡尔积
    SELECT DISTINCT m1.meeting,m2.person FROM Meetings AS m1 CROSS JOIN Meetings AS m2;
    -- 求出缺席者的SQL语句(1):存在量化的应用
    SELECT DISTINCT m1.meeting,m2.person 
    FROM Meetings AS m1 CROSS JOIN Meetings AS m2
    WHERE NOT EXISTS (SELECT * FROM Meetings AS m3 WHERE m1.meeting = m3.meeting AND m2.person = m3.person);
    -- 求出缺席者的SQL语句(2):使用差集运算
    SELECT m1.meeting,m2.person
    FROM Meetings AS m1,Meetings AS m2
    EXCEPT 
    SELECT meeting,person
    FROM Meetings;

    全称量词(1):习惯"肯定 \(\Leftrightarrow\) 双重否定"之间的转换

    /* 全称量化(1):习惯“肯定<=>双重否定”之间的转换 */
    CREATE TABLE TestScores
    (student_id INTEGER,
     subject    VARCHAR(32) ,
     score      INTEGER,
      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',90);
    INSERT INTO TestScores VALUES(300, '社会',55);
    INSERT INTO TestScores VALUES(400, '数学',80);
    -- 查出所有科目分数都在50分以上的学生
    SELECT DISTINCT student_id FROM TestScores AS TS1
    WHERE NOT EXISTS (SELECT * FROM TestScores AS TS2 WHERE TS2.student_id 
    = TS1.student_id AND TS2.score < 50) -- 仅就本题而言,还可以使用min(score)>=50
    -- 查找出数学分数>=80,语文分数>=50的学生
    SELECT DISTINCT student_id FROM TestScores AS TS1
    WHERE subject IN ('数学','语文') AND NOT EXISTS (SELECT * FROM TestScores AS TS2 WHERE TS2.student_id 
    = TS1.student_id AND 1 = CASE WHEN subject = '数学' AND score < 80 THEN 1
                                  WHEN subject = '语文' AND score < 50 THEN 1 ELSE 0 END)
    GROUP BY student_id
    HAVING COUNT(*) = 2; -- group by having 子句要求两门课程都要有成绩

    全称量化(2):集合和谓词,哪个更强大?

    /* 全称量化(2):集合VS谓词——哪个更强大? */
    CREATE TABLE Projects
    (project_id VARCHAR(32),
     step_nbr   INTEGER ,
     status     VARCHAR(32),
      PRIMARY KEY(project_id, step_nbr));
    
    INSERT INTO Projects VALUES('AA100', 0, '完成');
    INSERT INTO Projects VALUES('AA100', 1, '等待');
    INSERT INTO Projects VALUES('AA100', 2, '等待');
    INSERT INTO Projects VALUES('B200',  0, '等待');
    INSERT INTO Projects VALUES('B200',  1, '等待');
    INSERT INTO Projects VALUES('CS300', 0, '完成');
    INSERT INTO Projects VALUES('CS300', 1, '完成');
    INSERT INTO Projects VALUES('CS300', 2, '等待');
    INSERT INTO Projects VALUES('CS300', 3, '等待');
    INSERT INTO Projects VALUES('DY400', 0, '完成');
    INSERT INTO Projects VALUES('DY400', 1, '完成');
    INSERT INTO Projects VALUES('DY400', 2, '完成');
    -- 查询完成到了工程1的项目 having子句解法
    SELECT product_id FROM Projects GROUP BY project_id HAVING COUNT(*) = SUM(CASE WHEN step_nbr <= AND status = '完成' THEN 1 WHEN step_nbr > 1 AND status = '等待' THEN 1 ELSE 0 END);
    -- 查询完成到了工程1的项目 having子句解法 谓词解法
    SELECT * FROM Projects P1 WHERE NOT EXISTS (SELECT status from Projects P2 WHERE P1.project_id = P2.project_id AND status <> CASE WHEN step_nbr <= 1 THEN '完成' ELSE '等待' END);
    -- 劣势:双重否定,不易理解;优势:性能好,只要有一个行满足条件,查询就会终止;包含的信息更全

    对列进行量化:查询全是1的行

    /* 对列进行量化:查询全是1的行 */
    CREATE TABLE ArrayTbl
     (keycol CHAR(1) PRIMARY KEY,
      col1  INTEGER,
      col2  INTEGER,
      col3  INTEGER,
      col4  INTEGER,
      col5  INTEGER,
      col6  INTEGER,
      col7  INTEGER,
      col8  INTEGER,
      col9  INTEGER,
      col10 INTEGER);
    
    --全为NULL
    INSERT INTO ArrayTbl VALUES('A', NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL);
    INSERT INTO ArrayTbl VALUES('B', 3, NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL, NULL);
    --全为1
    INSERT INTO ArrayTbl VALUES('C', 1, 1, 1, 1, 1, 1, 1, 1, 1, 1);
    --至少有一个9
    INSERT INTO ArrayTbl VALUES('D', NULL, NULL, 9, NULL, NULL, NULL, NULL, NULL, NULL, NULL);
    INSERT INTO ArrayTbl VALUES('E', NULL, 3, NULL, 1, 9, NULL, NULL, 9, NULL, NULL);
    -- "列方向"的全称量化:查找全是1的行 不优雅的解答
    SELECT * FROM ArrayTbl WHERE col1 = 1 AND col2 = 1 AND col3 = 1 AND col4 = 1 AND col5 = 1
    AND col6 = 1 AND col7 = 1 AND col8 = 1 AND col9 = 1 AND col10 = 1;
    
    -- "列方向"的全称量化:查找全是1的行 优雅的解答
    SELECT * FROM ArrayTbl WHERE 1 = ALL (values(col1),(col2),(col3),(col4),(col5),(col6),(col7),(col8),(col9),(col10));
    
    -- "列方向"的全称量化:查找某一列是9的行
    SELECT * FROM ArrayTbl WHERE 9 = ANY (values(col1),(col2),(col3),(col4),(col5),(col6),(col7),(col8),(col9),(col10));
    
    -- "列方向"的全称量化:查找某一列是9的行
    SELECT * FROM ArrayTbl WHERE 9 IN (col1,col2,col3,col4,col5,col6,col7,col8,col9,col10);
    
    -- 查找全是NULL的行
    SELECT * FROM ArrayTbl WHERE COALESCE(col1,col2,col3,col4,col5,col6,col7,col8,col9,col10) IS NULL;

    小结

    • SQL中的谓词指的是返回真值的函数
    • EXISTS与其他谓词不同,接受的参数是集合
    • 因此EXISTS可以看成一种高阶函数
    • SQL没有与全称量词相当的谓词,可以使用NOT EXISTS代替

    练习题

    /* 练习题1-8-1:数组表——行结构表的情况 */
    CREATE TABLE ArrayTbl2
     (key   CHAR(1) NOT NULL,
        i   INTEGER NOT NULL,
      val   INTEGER,
      PRIMARY KEY (key, i));
    
    /* A全为NULL、B仅有一个为非NULL、C全为非NULL */
    INSERT INTO ArrayTbl2 VALUES('A', 1, NULL);
    INSERT INTO ArrayTbl2 VALUES('A', 2, NULL);
    INSERT INTO ArrayTbl2 VALUES('A', 3, NULL);
    INSERT INTO ArrayTbl2 VALUES('A', 4, NULL);
    INSERT INTO ArrayTbl2 VALUES('A', 5, NULL);
    INSERT INTO ArrayTbl2 VALUES('A', 6, NULL);
    INSERT INTO ArrayTbl2 VALUES('A', 7, NULL);
    INSERT INTO ArrayTbl2 VALUES('A', 8, NULL);
    INSERT INTO ArrayTbl2 VALUES('A', 9, NULL);
    INSERT INTO ArrayTbl2 VALUES('A',10, NULL);
    INSERT INTO ArrayTbl2 VALUES('B', 1, 3);
    INSERT INTO ArrayTbl2 VALUES('B', 2, NULL);
    INSERT INTO ArrayTbl2 VALUES('B', 3, NULL);
    INSERT INTO ArrayTbl2 VALUES('B', 4, NULL);
    INSERT INTO ArrayTbl2 VALUES('B', 5, NULL);
    INSERT INTO ArrayTbl2 VALUES('B', 6, NULL);
    INSERT INTO ArrayTbl2 VALUES('B', 7, NULL);
    INSERT INTO ArrayTbl2 VALUES('B', 8, NULL);
    INSERT INTO ArrayTbl2 VALUES('B', 9, NULL);
    INSERT INTO ArrayTbl2 VALUES('B',10, NULL);
    INSERT INTO ArrayTbl2 VALUES('C', 1, 1);
    INSERT INTO ArrayTbl2 VALUES('C', 2, 1);
    INSERT INTO ArrayTbl2 VALUES('C', 3, 1);
    INSERT INTO ArrayTbl2 VALUES('C', 4, 1);
    INSERT INTO ArrayTbl2 VALUES('C', 5, 1);
    INSERT INTO ArrayTbl2 VALUES('C', 6, 1);
    INSERT INTO ArrayTbl2 VALUES('C', 7, 1);
    INSERT INTO ArrayTbl2 VALUES('C', 8, 1);
    INSERT INTO ArrayTbl2 VALUES('C', 9, 1);
    INSERT INTO ArrayTbl2 VALUES('C',10, 1);
    /* 正确解法 */
    SELECT DISTINCT key
      FROM ArrayTbl2 A1
     WHERE NOT EXISTS
            (SELECT *
               FROM ArrayTbl2 A2
              WHERE A1.key = A2.key
                AND (A2.val <> 1 OR A2.val IS NULL));
    /* 其他解法1:使用ALL谓词 */
    SELECT DISTINCT key
      FROM ArrayTbl2 A1
     WHERE 1 = ALL
              (SELECT val
                 FROM ArrayTbl2 A2
                WHERE A1.key = A2.key);
    /* 其他解法2:使用HAVING子句 */
    SELECT key
      FROM ArrayTbl2
     GROUP BY key
    HAVING SUM(CASE WHEN val = 1 THEN 1 ELSE 0 END) = 10;
    /* 其他解法3:在HAVING子句中使用极值函数 */
    SELECT key
      FROM ArrayTbl2
     GROUP BY key
    HAVING MAX(val) = 1
       AND MIN(val) = 1;
    /* 练习题1-8-2:使用ALL谓词进行全称量化
       查找已经完成到工程1的项目:使用ALL谓词解答 */
    SELECT *
      FROM Projects P1
     WHERE '○' = ALL
                 (SELECT CASE WHEN step_nbr <= 1 AND status = '完成' THEN '○'
                              WHEN step_nbr > 1  AND status = '等待' THEN '○'
                              ELSE '×' END
                    FROM Projects P2
                   WHERE P1.project_id = P2. project_id);
    
    /* 练习题1-8-3:求(1-100)中的质数 */
    SELECT num AS prime
      FROM Numbers Dividend
     WHERE num > 1
       AND NOT EXISTS
            (SELECT *
               FROM Numbers Divisor
              WHERE Divisor.num <= Dividend.num / 2 /* 除了自身之外的约数必定小于等于自身值的一半 */
                AND Divisor.num <> 1 /* 约数中不包含1 */
                AND MOD(Dividend.num, Divisor.num) = 0)  /*“除不尽”的否定条件是“除尽” */
    ORDER BY prime;

    SQL进阶系列之8EXISTS谓词的用法

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