给定一组课程，我需要确定参加每个课程的学生所需的房间总数和类型。课程可以并行运行，嵌套运行或重叠运行。
要实现的逻辑：对于每个课程持续时间，找到在该持续时间内所有其他活动课程，以及Number_of_students总数
这些课程中有哪些按room_type分组。
还存在其他问题，但下面给出了该问题的简化版本。
我目前使用hsqldb，该解决方案应使用std sql语法在数据库之间移植。
预订表
BOOKING_ID| COURSE_ID| NUMBER_OF_STUDENTS| ROOM_TYPE_ID
10 | 2 | 1 | 1
20 | 1 | 2 | 1
30 | 3 | 1 | 3
40 | 1 | 3 | 4
50 | 5 | 1 | 2
60 | 6 | 2 | 2
70 | 3 | 2 | 1
80 | 4 | 1 | 3
课程表
COURSE_ID| START_DATE | END_DATE
1 | 2018-05-15 | 2018-06-14 //sample course
2 | 2018-05-11 | 2018-05-20 //starts before ends between sample course
3 | 2018-05-18 | 2018-05-22 //starts between ends between sample course
4 | 2018-05-20 | 2018-06-20 //starts between ends after sample course
5 | 2018-05-10 | 2018-06-20 //starts before ends after sample course
6 | 2018-05-10 | 2018-05-14 //starts and ends before sample course
7 | 2018-06-15 | 2018-06-20 //starts and ends after sample course
Rooms Table（出于完整性考虑，我们在这里确实不需要此表）
ROOM_TYPE_ID| ROOM_CAPACITY| ROOM_LOCATION
1 | 1 | HILL
2 | 2 | HILL
3 | 1 | OCEAN
4 | 2 | OCEAN
输出（仅对course_id 1显示，所有课程均必需）
COURSE_ID | ROOMTYPE | COURSE_STUDENT | OTHER_STUDENTS
1 | 1 | 2 | 3 //1(course 2) + 2 (course 3)
1 | 2 | 0 | 1 //1(course 5)
1 | 3 | 0 | 2 //1(course 3) + 1(course 4)
1 | 4 | 3 | 0 //no students on others
我只能找出给定课程 startDate ， endDate 匹配重叠课程的条件
Courses.START_DATE <= startDate AND Courses.END_DATE >= endDate OR //matches any course spanning current course
Courses.START_DATE >= startDate AND Courses.END_DATE <= startDate OR //matches any course starting during the current course
Courses.START_DATE >= endDate AND Courses.END_DATE <= endDate //matches any course ending during the current course
除了我微薄的sql技能之外，我还惨遭失败。我可以旋转一些Java代码来解决此问题。...但是那简直是la脚&&低效。
答案 0 :(得分：1)
感谢弗雷德（Fredt）向我指出了正确的方向...
将主表及其所有记录均连接到所有记录，然后根据重叠的课程日期条件进行过滤。
下面的查询效率不高，但目前可以完成工作。...可能还有更多优化，我想听听其他意见
SELECT
THIS_COURSE.COURSE_ID,
OTHER_COURSE.ROOM_TYPE_ID,
SUM(CASE WHEN THIS_COURSE.BOOKING_ID = OTHER_COURSE.BOOKING_ID THEN OTHER_COURSE.NUMBER_OF_STUDENTS ELSE 0 END) AS COURSE_STUDENTS,
SUM(CASE WHEN THIS_COURSE.BOOKING_ID <> OTHER_COURSE.BOOKING_ID THEN OTHER_COURSE.NUMBER_OF_STUDENTS ELSE 0 END) AS OTHER_STUDENTS,
SUM(OTHER_COURSE.NUMBER_OF_STUDENTS) AS TOTAL_STUDENTS
FROM
(
SELECT
BOOKINGS.BOOKING_ID,
BOOKINGS.COURSE_ID,
BOOKINGS.NUMBER_OF_STUDENTS,
BOOKINGS.ROOM_TYPE_ID,
COURSES.START_DATE,
COURSES.END_DATE
FROM
BOOKINGS , COURSES
WHERE
BOOKINGS.COURSE_ID = COURSES.COURSE_ID
) THIS_COURSE
LEFT JOIN
(
SELECT
BOOKINGS.BOOKING_ID,
BOOKINGS.COURSE_ID,
BOOKINGS.NUMBER_OF_STUDENTS,
BOOKINGS.ROOM_TYPE_ID,
COURSES.START_DATE,
COURSES.END_DATE
FROM
BOOKINGS , COURSES
WHERE
BOOKINGS.COURSE_ID = COURSES.COURSE_ID
) OTHER_COURSE
ON
THIS_COURSE.BOOKING_ID <> OTHER_COURSE.BOOKING_ID OR
THIS_COURSE.BOOKING_ID = OTHER_COURSE.BOOKING_ID
WHERE
(THIS_COURSE.START_DATE <= OTHER_COURSE.START_DATE AND THIS_COURSE.END_DATE >= OTHER_COURSE.END_DATE) OR
(THIS_COURSE.START_DATE <= OTHER_COURSE.START_DATE AND THIS_COURSE.END_DATE >= OTHER_COURSE.START_DATE) OR
(THIS_COURSE.START_DATE <= OTHER_COURSE.END_DATE AND THIS_COURSE.END_DATE >= OTHER_COURSE.END_DATE)
GROUP BY
THIS_COURSE.COURSE_ID, OTHER_COURSE.ROOM_TYPE_ID
下面是创建示例数据的sql
CREATE TABLE Bookings(BOOKING_ID INTEGER NOT NULL PRIMARY KEY, COURSE_ID INTEGER NOT NULL, NUMBER_OF_STUDENTS INTEGER NOT NULL, ROOM_TYPE_ID INTEGER NOT NULL)
CREATE TABLE Courses(COURSE_ID INTEGER NOT NULL PRIMARY KEY, START_DATE DATE, END_DATE DATE)
CREATE TABLE Rooms(ROOM_TYPE_ID INTEGER NOT NULL PRIMARY KEY, ROOM_CAPACITY INTEGER NOT NULL, ROOM_LOCATION VARCHAR(25))
INSERT INTO Bookings VALUES( 10 , 2 , 1 , 1 )
INSERT INTO Bookings VALUES( 20 , 1 , 2 , 1 )
INSERT INTO Bookings VALUES( 30 , 3 , 1 , 3 )
INSERT INTO Bookings VALUES( 40 , 1 , 3 , 4 )
INSERT INTO Bookings VALUES( 50 , 5 , 1 , 2 )
INSERT INTO Bookings VALUES( 60 , 6 , 2 , 2 )
INSERT INTO Bookings VALUES( 70 , 3 , 2 , 1 )
INSERT INTO Bookings VALUES( 80 , 4 , 1 , 3 )
INSERT INTO Bookings VALUES( 90 , 7 , 1 , 4 )
INSERT INTO Courses VALUES( 1 ,'2018-05-15', '2018-06-14' )
INSERT INTO Courses VALUES( 2 ,'2018-05-11', '2018-05-20' )
INSERT INTO Courses VALUES( 3 ,'2018-05-18', '2018-05-22' )
INSERT INTO Courses VALUES( 4 ,'2018-05-20', '2018-06-20' )
INSERT INTO Courses VALUES( 5 ,'2018-05-10', '2018-06-20' )
INSERT INTO Courses VALUES( 6 ,'2018-05-10', '2018-05-14' )
INSERT INTO Courses VALUES( 7 ,'2018-06-15', '2018-06-20' )
INSERT INTO Rooms VALUES( 1 , 1 , 'HILL')
INSERT INTO Rooms VALUES( 2 , 2 , 'HILL')
INSERT INTO Rooms VALUES( 3 , 1 , 'OCEAN')
INSERT INTO Rooms VALUES( 4 , 2 , 'OCEAN')
答案 1 :(得分：0)
您实际上想要每种课程每种房间所需的房间数量。因此，您需要从COURSES表开始并将其与其他两个表连接起来。
SELECT * FROM COURSES JOIN BOOKINGS USING (COURSE_ID) JOIN ROOMS USING (ROOM_TYPE_ID)
这为您提供了所有客房预订的一长串清单。然后，您可以将此表视为子查询表，并根据日期时间段将其自身联接起来。
WITH ROOM_BOOKINGS AS (
SELECT
BOOKINGS.BOOKING_ID,
BOOKINGS.COURSE_ID,
BOOKINGS.NUMBER_OF_STUDENTS,
BOOKINGS.ROOM_TYPE_ID,
COURSES.START_DATE,
COURSES.END_DATE,
ROOMS.ROOM_CAPACITY
FROM
COURSES JOIN BOOKINGS USING (COURSE_ID) JOIN ROOMS USING (ROOM_TYPE_ID)
)
SELECT * FROM ROOM_BOOKINGS THIS_COURSE LEFT JOIN ROOM_BOOKINGS OTHER_COURSE
ON (THIS_COURSE.START_DATE, THIS_COURSE.END_DATE + 1 DAY) OVERLAPS (OTHER_COURSE.START_DATE, OTHER_COURSE.END_DATE + 1 DAY)
AND THIS_COURSE.ROOM_TYPE_ID = OTHER_COURSE.ROOM_TYPE_ID
AND THIS_COURSE.COURSE_ID <> OTHER_COURSE.COURSE_ID
您需要完成上述查询并将条件添加到SELECT中，以仅返回一门课程。您还需要GROUP BY A.COURSE_ID，A.ROOM_TYPE_ID，A.NUMBER_OF_STUDENTS，...和SUM（B.NUMBER_OF_STUDENS）才能获得所需的输出。
如您所见，编写高级SQL查询并不是一项繁琐的任务，并且需要对SQL语言有充分的了解。