Question

对于模糊的标题感到抱歉（我只是不知道如何描述这个难题）

为教室提供以下计划表：

╔═══════════╦════════════╦═══════════╦═══════════╦═════════╗
║ Classroom ║ CourseName ║  Lesson   ║ StartTime ║ EndTime ║
╠═══════════╬════════════╬═══════════╬═══════════╬═════════╣
║      1001 ║ Course 1   ║ Lesson 1  ║      0800 ║    0900 ║
║      1001 ║ Course 1   ║ Lesson 2  ║      0900 ║    1000 ║
║      1001 ║ Course 1   ║ Lesson 3  ║      1000 ║    1100 ║
║      1001 ║ Course 2   ║ Lesson 10 ║      1100 ║    1200 ║
║      1001 ║ Course 2   ║ Lesson 11 ║      1200 ║    1300 ║
║      1001 ║ Course 1   ║ Lesson 4  ║      1300 ║    1400 ║
║      1001 ║ Course 1   ║ Lesson 5  ║      1400 ║    1500 ║
╚═══════════╩════════════╩═══════════╩═══════════╩═════════╝

我想将表格分组以显示：

╔═══════════╦════════════╦═══════════╦═════════╗
║ Classroom ║ CourseName ║ StartTime ║ EndTime ║
╠═══════════╬════════════╬═══════════╬═════════╣
║      1001 ║ Course 1   ║      0800 ║    1100 ║
║      1001 ║ Course 2   ║      1100 ║    1300 ║
║      1001 ║ Course 1   ║      1300 ║    1500 ║
╚═══════════╩════════════╩═══════════╩═════════╝

基本上我们正在查看一个时间表，显示在某个时间段内哪个教室正在使用什么教室......

我最初的想法是：按Classroom和CourseName分组，Max和Min start\end Classroom时间，但这不会给我时间跨度，就像课程1一样使用{{1}}从08:00到16:00，中间没有休息。

Answer 1

查询使用EndTime确定每一行NOT EXISTS，以确保在课程范围StartTime和EndTime之间没有安排其他类别或课程。然后使用MIN和GROUP BY查找StartTime。

NOT EXISTS部分通过搜索StartTime之间EndTime的任何行，确保EndTime和StartTime范围之间没有“中断” }和EndTime但属于不同的CourseName或CourseRoom。

SELECT    
    t0.ClassRoom,
    t0.CourseName,
    MIN(t0.StartTime),
    t0.EndTime
FROM (
    SELECT 
    t1.ClassRoom,
    t1.CourseName,
    t1.StartTime,
    (
        SELECT MAX(t2.EndTime)
        FROM tableA t2
        WHERE t2.CourseName = t1.CourseName
        AND t2.ClassRoom = t1.ClassRoom
        AND NOT EXISTS (SELECT 1 FROM tableA t3
            WHERE t3.EndTime < t2.EndTime 
            AND t3.EndTime > t1.EndTime
            AND (t3.CourseName <> t2.CourseName 
            OR t3.ClassRoom <> t2.ClassRoom)
        )
    ) EndTime
    FROM tableA t1
) t0 GROUP BY t0.ClassRoom, t0.CourseName, t0.EndTime

http://www.sqlfiddle.com/#!6/39d4b/9

Answer 2

如果您正在使用SQLServer 2012或更高版本，则可以使用LAG获取列的先前值，然后使用SUM() OVER (ORDER BY ...)创建滚动总和，在这种情况下，可以计算更改CourseName，可以用作GROUP BY锚点

With A AS (
  SELECT ClassRoom
       , CourseName
       , StartTime
       , EndTime
       , PrevCourse = LAG(CourseName, 1, CourseName) OVER (ORDER BY StartTime)
  FROM   Table1
), B AS (
  SELECT ClassRoom
       , CourseName
       , StartTime
       , EndTime
       , Ranker = SUM(CASE WHEN CourseName = PrevCourse THEN 0 ELSE 1 END)
                OVER (ORDER BY StartTime, CourseName)
  FROM   A
)
SELECT ClassRoom
     , CourseName
     , MIN(StartTime) StartTime
     , MAX(EndTime) EndTime
FROM   B
GROUP BY ClassRoom, CourseName, Ranker
ORDER BY StartTime

SQLFiddle demo

Answer 3

     CREATE TABLE Classroom(Classroom VARCHAR(100), CourseName  VARCHAR(100),  Lesson    VARCHAR(100), StartTime  VARCHAR(100), EndTime  VARCHAR(100))
 INSERT INTO Classroom
SELECT '1001','Course 1','Lesson 1 ','0800','0900'
UNION SELECT '1001','Course 1','Lesson 2 ','0900','1000'
UNION SELECT '1001','Course 1','Lesson 3 ','1000','1100'
UNION SELECT '1001','Course 2','Lesson 10','1100','1200'
UNION SELECT '1001','Course 2','Lesson 11','1200','1300'
UNION SELECT '1001','Course 1','Lesson 4 ','1300','1400'
UNION SELECT '1001','Course 1','Lesson 5 ','1400','1500'

SELECT * FROM Classroom

;WITH CTE_ClassRooms AS (
SELECT *,ROW_NUMBER() over(partition by classroom,CourseName order by StartTime) AS R FROM Classroom A 
WHERE NOT EXISTS(SELECT 1 FROM Classroom B WHERE B.Classroom = A.Classroom AND B.CourseName = A.CourseName AND B.StartTime = A.EndTime)
UNION ALL
SELECT B.*,R fROM CTE_ClassRooms A JOIN Classroom B ON B.Classroom = A.Classroom AND B.CourseName = A.CourseName AND A.StartTime = B.EndTime
)

--SELECT * FROM CTE_ClassRooms order by Classroom,CourseName,R

SELECT Classroom,CourseName,MIN(StartTime),MAX(EndTime)
FROM CTE_ClassRooms
GROUP BY Classroom,CourseName,R

使用时间跨度对相同值的连续行进行分组

3 个答案: