在SQL Server中查找时间范围内某些特殊项的计数

假设某人有一个任务，组织应该在任务期间支付他的餐费，但在任务结束之前不支付，例如，如果他在上午8点之后开始任务，他将不会得到任何报酬。 如果他在晚上9点后完成任务，也会是一样的，而不是晚餐时间，午餐时间在12点结束

这是桌子：

ID  Date        DepartureTime              breakfast  Lunch  Dinner
1   2018-03-13  2018-03-13 13:12:32.000    True       False  True
2   2018-03-14  NULL                       True       True   False
3   2018-03-15  NULL                       False      True   True
4   2018-03-16  NULL                       True       True   False
5   2018-03-17  2018-03-17 13:00:00.000    False      True   True

我需要一个查询，在这个查询中，我可以计算出符合支付条件的膳食数量

我想应该通过创建一个时间范围并在需要时使用CASE来解决这个问题

这里，True表示他吃了那顿饭，false表示他没有吃东西。 答案应该是：

---

早餐：2份；午餐：4天；晚餐：2

1将每顿饭与时间间隔或确切时间联系起来。例如：

早餐：上午8点至10点 午餐：从上午12点到下午2点 晚餐时间：晚上7点至10点 示例如下：

IF OBJECT_ID('tempdb..#MealTime') IS NOT NULL
    DROP TABLE #MealTime

CREATE TABLE #MealTime (
    Meal VARCHAR(100),
    StartTime TIME,
    EndTime TIME)

INSERT INTO #MealTime (
    Meal,
    StartTime,
    EndTime)
SELECT
    Meal = 'Breakfast',
    StartTime = '08:00',
    EndTime = '10:00'
UNION ALL
SELECT
    Meal = 'Lunch',
    StartTime = '12:00',
    EndTime = '14:00'
UNION ALL
SELECT
    Meal = 'Dinner',
    StartTime = '19:00',
    EndTime = '22:00'

2确定每行的出发时间表示开始还是结束，并创建时间间隔以简化步骤3：

;WITH Limits AS
(
    SELECT
        MinimumID = MIN(M.ID),
        MaximumID = MAX(M.ID)
    FROM
        OperationDetails AS M
)
SELECT
    M.*,
    StartTime = CONVERT(TIME,
        CASE 
            WHEN L1.MinimumID IS NOT NULL THEN M.DepartureTime
            ELSE '00:00' END),
    EndTime = CONVERT(TIME,
        CASE 
            WHEN L2.MaximumID IS NOT NULL THEN M.DepartureTime
            ELSE '23:59' END)
FROM
    OperationDetails AS M
    LEFT JOIN Limits AS L1 ON M.ID = L1.MinimumID
    LEFT JOIN Limits AS L2 ON M.ID = L2.MaximumID

3如果每顿饭都吃了，且时间重叠，则计算每顿饭的数量：

;WITH Limits AS
(
    SELECT
        MinimumID = MIN(M.ID),
        MaximumID = MAX(M.ID)
    FROM
        OperationDetails AS M
),
MissionTimes AS
(
    SELECT
        M.*,
        StartTime = CONVERT(TIME,
            CASE 
                WHEN L1.MinimumID IS NOT NULL THEN M.DepartureTime
                ELSE '00:00' END),
        EndTime = CONVERT(TIME,
            CASE 
                WHEN L2.MaximumID IS NOT NULL THEN M.DepartureTime
                ELSE '23:59' END)
    FROM
        OperationDetails AS M
        LEFT JOIN Limits AS L1 ON M.ID = L1.MinimumID
        LEFT JOIN Limits AS L2 ON M.ID = L2.MaximumID
)
SELECT
    AmountBreakfastToPay = COUNT(CASE WHEN M.Breakfast = 'true' AND M.StartTime <= BR.EndTime AND M.EndTime >= BR.StartTime THEN 1 END),
    AmountLunchToPay = COUNT(CASE WHEN M.Lunch = 'true' AND M.StartTime <= LU.EndTime AND M.EndTime >= LU.StartTime THEN 1 END),
    AmountDinnerToPay = COUNT(CASE WHEN M.Dinner = 'true' AND M.StartTime <= DI.EndTime AND M.EndTime >= DI.StartTime THEN 1 END)
FROM
    MissionTimes AS M
    INNER JOIN #MealTime AS BR ON BR.Meal = 'Breakfast'
    INNER JOIN #MealTime AS LU ON LU.Meal = 'Lunch'
    INNER JOIN #MealTime AS DI ON DI.Meal = 'Dinner'

---

有了这些样本数据，预期的结果是什么？你能告诉我们你当前的查询尝试吗？对我来说毫无意义。当你不能解决它的时候，你为什么要投反对票，我已经解释得够多了，当你没有资格解决它的时候，就别管它了！！！出发时间是否意味着任务的开始时间ID=1和结束时间ID=5？为什么出发时间与ID 5上的日期不匹配？不幸的是，是的，任务a的开始时间和结束时间由出发时间规定，是他离开家或完成任务的时间。

SELECT *
,B=CASE WHEN (cast(DepartureTime AS time) between '00:00:00' AND '08:00:00' OR DepartureTime IS NULL) AND [Breakfast]='True'  THEN 1 END
,L=CASE WHEN (cast(DepartureTime AS time) between '08:00:01' AND '13:00:00' OR DepartureTime IS NULL) AND [Lunch]='True'  THEN 1 END
,D=CASE WHEN (cast(DepartureTime AS time) between '13:00:01' AND '20:00:00' OR DepartureTime IS NULL) AND [Dinner]='True'  THEN 1 END
FROM MissionToEat