MySQL：用 SQL 找出日期的工作日是否在两个工作日之间

Question

我有这个 vb.NET 代码，想在 MySQL 中写一个函数来做同样的事情。

Public Function IsWeekdayTimeInRange(DateToCheck As DateTime, StartDayOfWeek As DayOfWeek,
                                                              EndDayOfWeek As DayOfWeek) As Boolean

    Dim WeekdaysInRange As New List(Of DayOfWeek)
    WeekdaysInRange.Add(StartDayOfWeek)
    Dim i As Integer = StartDayOfWeek
    While i <> EndDayOfWeek
        i += 1
        i = If(i = 7, 0, i)
        WeekdaysInRange.Add(i)
    End While

    Return WeekdaysInRange.Contains(DateToCheck.DayOfWeek)
End Function

我目前在 MySQL 中，但由于我从未编写过 SQL 函数，因此我需要一些帮助。

DELIMITER |
CREATE FUNCTION IS_BETWEEN_WEEKDAYS(d DATETIME, s INT, e INT)
RETURNS BOOLEAN DETERMINISTIC
BEGIN
    DECLARE weekdayToCheck INT;
    DECLARE result BOOLEAN;
    DECLARE i INT;
    SET result = false;
    SET weekdayToCheck = WEEKDAY(d);
    SET i = s;
    WHILE(i <> e) DO

        /* more logic here */

    END WHILE;

    RETURN result;
END; |

DELIMITER ;

编辑：由于我现在了解到 MySQL 具有 Weekday(Monday) = 0，因此我的函数也需要转换为 Weekday(Sunday) = 0，因为我的数据使用该格式。

Answer 1

现在可以了。它可能更简洁，但它帮助我思考。

DELIMITER |
CREATE FUNCTION IS_BETWEEN_WEEKDAYS(d DATETIME, s INT, e INT, FirstDayOfWeekIsSunday BOOLEAN)
RETURNS BOOLEAN DETERMINISTIC
BEGIN
    DECLARE weekdayToCheck INT;
    DECLARE result BOOLEAN;
    DECLARE i INT;
    SET result = false;
    SET weekdayToCheck = WEEKDAY(d);
    SET i = s;

    IF FirstDayOfWeekIsSunday = TRUE THEN
        SET weekdayToCheck = weekdayToCheck + 1;
        IF weekdayToCheck = 7 THEN
            SET weekdayToCheck = 0;
        END IF;
    END IF;

    IF weekdayToCheck = s THEN
        SET result = true;
    END IF;

    WHILE(i <> e) DO

        SET i = i + 1;

        IF i = 7 THEN
            SET i = 0;
        END IF;

        IF weekdayToCheck = i THEN 
            SET result = true;
        END IF;

    END WHILE;

    RETURN result;
END; |

DELIMITER ;

编辑：修复了循环中的顺序问题。还添加了处理s 和e 的选项，如果他们假设一周从星期日开始。 编辑：测试了上述内容并且可以正常工作。还可以创建一个函数来检查日期时间是否在星期二下午 3 点和星期四凌晨 2 点之间。

DELIMITER |
CREATE FUNCTION IS_BETWEEN_WEEKDAYS_AND_TIME(d DATETIME, s INT, starttime TIME, e INT, endtime TIME, FirstDayOfWeekIsSunday BOOLEAN)
RETURNS BOOLEAN DETERMINISTIC
BEGIN
    DECLARE weekdayToCheck INT;
    DECLARE result BOOLEAN;
    DECLARE i INT;
    DECLARE dTime TIME;
    SET result = false;
    SET weekdayToCheck = WEEKDAY(d);
    SET i = s;
    SET dTime = DATE_FORMAT(d, '%H:%i:%s');

    IF FirstDayOfWeekIsSunday = TRUE THEN
        SET weekdayToCheck = weekdayToCheck + 1;
        IF weekdayToCheck = 7 THEN
            SET weekdayToCheck = 0;
        END IF;
    END IF;

    IF weekdayToCheck = s THEN 
        IF ((weekdayToCheck = s AND dTime > starttime) OR (weekdayToCheck = e AND dTime < endtime) OR (weekdayToCheck <> s AND weekdayToCheck <> e)) THEN
            SET result = true;
        END IF;
    END IF;

    WHILE(i <> e) DO

        SET i = i + 1;

        IF i = 7 THEN
            SET i = 0;
        END IF;

        IF weekdayToCheck = i THEN 
            IF ((weekdayToCheck = s AND dTime > starttime) OR (weekdayToCheck = e AND dTime < endtime) OR (weekdayToCheck <> s AND weekdayToCheck <> e)) THEN
                SET result = true;
            END IF;
        END IF;

    END WHILE;

    RETURN result;
END; |

DELIMITER ;

Answer 2

考虑从星期一开始的一周。希望这会有所帮助：

DELIMITER |
CREATE FUNCTION IS_BETWEEN_WEEKDAYS(d DATETIME, s INT, e INT)
RETURNS BOOLEAN DETERMINISTIC
BEGIN
  DECLARE weekdayToCheck INT;
  DECLARE result TINYINT(1);
  SET weekdayToCheck =  WEEKDAY(d);
  SET result = 0;
  IF(weekdayToCheck > s AND weekdayToCheck < e) THEN
    SET result = 1;
  ELSEIF (s > e AND weekdayToCheck < s AND weekdayToCheck < e) THEN
    SET result = 1;
  END IF;

RETURN result;
END; |

DELIMITER ;

Answer 3

不确定你对“之间”的定义是什么，但在这里我使用了大于和小于

CREATE FUNCTION IS_BETWEEN_WEEKDAYS(d DATETIME, s INT, e INT)
RETURNS BOOLEAN DETERMINISTIC
BEGIN
    DECLARE weekdayToCheck INT;
    DECLARE result INT;
    SET weekdayToCheck = WEEKDAY(d);
    SET result = 0;

    IF (s < e) THEN 
       IF  (weekdayToCheck > s AND weekdayToCheck < e) THEN SET result = 1;
       END IF;
    END IF;

    IF (s > e) THEN
       IF (weekdayToCheck > s OR weekdayToCheck < e) THEN SET result = 1;
       END IF;        
    END IF;

    RETURN result; 
END;

也许所有的 if/else 子句都可以缩短一些。