Question

我有用于管理体育学院的数据库。现在我有玩家，类和class_player的表（加入其他两个表）。我想要做的是当将玩家注册到classA时，注册到classB的玩家（与classA时间相交）将从select语句的结果集中删除

Player Table:         class Table               class_player Table
ID INT                ID INT                    classID INT
                      timeFrom TIME             playerID INT
                      timeTo TIME
                      days BIT(7)

In＆＃34; class＆＃34;表： timeFrom：是上课时间的开始 timeTo：是上课时间的结束 days：掌握上课的日子，
例如：0x000001 - ＆gt;七天中的一天将包括这个课程

我用来获取玩家的选择语句：

(
SELECT `players`.`ID`
FROM `players` JOIN `class_player` ON `players`.`ID` = `class_player`.`playerID`
JOIN `classes` ON `class_player`.`classID` = `classes`.`ID`
WHERE (
    -- Get all player with no day intersection
    -- 1 is a variable of the entered day ex: 0x00100100
    NOT (`classes`.`days` & 1)

    -- Time Intersection
    OR (
            -- Entered "FROM" != Existing "FROM"
            '10:00:00' != `timeFrom` 

            -- Entered "TO" != Existing "TO"
            AND '13:00:00' != `timeTo`

            -- Entered "FROM" Not Between Existing (FROM, TO)
            AND NOT ('10:00:00' > `timeFrom` AND  '10:00:00' < `timeTo`)
            -- AND '10:00:00' NOT BETWEEN `timeFrom` AND `timeTo`

            -- Entered "TO" Not Between Existing (FROM, TO)
            AND NOT ('13:00:00' < `timeTo` AND '13:00:00' > `timeFrom`)
            -- AND '13:00:00' NOT BETWEEN `timeFrom` AND `timeTo` 

            -- Entered "FROM" Not Less than Existing "FROM"
            -- Entered "TO" Not Bigger than Existing "TO"
            AND NOT ('10:00:00' < `timeFrom` AND  '13:00:00' > `timeTo` )

        )
    )
)
UNION
(
    -- Players who don't have any classes
    SELECT `players`.`ID`
    FROM `players`
    WHERE `players`.`ID` NOT IN (SELECT `playerID` FROM `class_player`)
);

问题是当一个玩家有两个班级，一个班级与班级时间不相交时，这个玩家会在结果集中显示!!

Answer 1

首先：感谢wojciechz他的努力，解决方案是他的第二个答案，但我不理解，直到我的朋友告诉我类似的解决方案。

第二：SQL中的解决方案是：
1.让所有忙碌的球员得到 2.然后搜索不在此列表中的玩家

SELECT DISTINCT `ID` FROM `players` WHERE `ID` NOT IN (
    (
        SELECT `players`.`ID`
        FROM `players` JOIN `class_player` ON `players`.`ID` = `class_player`.`playerID`
        JOIN `classes` ON `class_player`.`classID` = `classes`.`ID`
        WHERE (
            -- Get all player with no day intersection
            -- 1 is a variable of the entered day ex: 0x00100100
            (`classes`.`days` & 1)

            -- Time Intersection
            AND (
                    -- Entered "FROM" != Existed "FROM"
                    '10:00:00' = `timeFrom` 

                    -- Entered "TO" != Existed "TO"
                    OR '13:00:00' = `timeTo`

                    -- Entered "FROM" Not Between Existing (FROM, TO)
                    OR ('10:00:00' > `timeFrom` AND  '10:00:00' < `timeTo`)
                    -- AND '10:00:00' NOT BETWEEN `timeFrom` AND `timeTo`

                    -- Entered "TO" Not Between Existing (FROM, TO)
                    OR ('13:00:00' < `timeTo` AND '13:00:00' > `timeFrom`)
                    -- AND '13:00:00' NOT BETWEEN `timeFrom` AND `timeTo` 

                    -- Entered "FROM" Not Less than Existing "FROM"
                    -- Entered "TO" Not Bigger than Existing "TO"
                    OR ('10:00:00' < `timeFrom` AND  '13:00:00' > `timeTo` )

            )
        )
    )
    UNION
    (
        -- ANOTHER DATE AND TIME
    )
)

如何消除MySQL中不符合某些条件的行

1 个答案: