我有以下数据：
CREATE TABLE Table1
(
ID varchar(10),
StudentName varchar(30),
Course varchar(15),
SECTION varchar(2),
DAY varchar(10),
START_TIME time,
END_TIME time,
actual_starttime time,
actual_endtime time
);
INSERT INTO Table1
VALUES (111, 'Mary', 'Science', 'A', 'Mon', '13:30:00.0000000', '16:20:00.0000000', '09:00:00.0000000', '21:20:00.0000000')
INSERT INTO Table1
VALUES (111, 'Mary', 'Maths', 'A', 'Tue', '12:30:00.0000000', '13:20:00.0000000', '09:00:00.0000000', '21:20:00.0000000')
INSERT INTO Table1
VALUES (111, 'Mary', 'Physics', 'C', 'Tue', '10:30:00.0000000', '11:10:00.0000000', '09:00:00.0000000', '21:20:00.0000000')
INSERT INTO Table1
VALUES (112, 'Robert', 'Maths', 'A', 'Mon', '13:30:00.0000000', '16:20:00.0000000', '09:00:00.0000000', '21:20:00.0000000')
场景如下：学生可以在周一至周五的早上9点到晚上9点30分。我的要求是我必须确定一个时间段，同一部分的所有学生都是免费的，以便教师可以重新安排课程。
例子：玛丽和罗伯特在星期一下午9点到1点30分都是免费的。我想为此写一个查询。
请帮忙。
提前致谢！
答案 0 :(得分：0)
您可以使用以下步骤创建存储过程。
步骤1：在另一个表格中预定义时间段（09：00-10：00,10：00-11：00等）
第2步：选择学生人数
第3步：
for all the slots
Begin
for all the students
Begin
if(students.actual_starttime =slots.actual_starttime and
students.actual_endtime =slots.actual_endtime
break;
else count=count+1;
End
End
步骤4：如果以上计数与学生总数相匹配，则所有学生都可以免费使用插槽。其他插槽不适合所有学生。
希望这会有所帮助。如果你发现它有困难，请告诉我。
答案 1 :(得分：0)
你应该再增加三个表来使它更简单 即学生，科和插槽
我尝试再创建一个半小时的桌子
create table table2(timeslot time);
insert into table2 values ('9:00:00.0000000');
insert into table2 values ('9:30:00.0000000');
insert into table2 values ('10:00:00.0000000');
insert into table2 values ('10:30:00.0000000');
insert into table2 values ('11:00:00.0000000');
insert into table2 values ('11:30:00.0000000');
insert into table2 values ('12:00:00.0000000');
insert into table2 values ('12:30:00.0000000');
insert into table2 values ('13:00:00.0000000');
insert into table2 values ('13:30:00.0000000');
insert into table2 values ('14:00:00.0000000');
insert into table2 values ('14:30:00.0000000');
insert into table2 values ('15:00:00.0000000');
insert into table2 values ('15:30:00.0000000');
insert into table2 values ('16:00:00.0000000');
insert into table2 values ('16:30:00.0000000');
insert into table2 values ('17:00:00.0000000');
insert into table2 values ('17:30:00.0000000');
insert into table2 values ('18:00:00.0000000');
insert into table2 values ('18:30:00.0000000');
insert into table2 values ('19:00:00.0000000');
insert into table2 values ('19:30:00.0000000');
insert into table2 values ('20:00:00.0000000');
insert into table2 values ('20:30:00.0000000');
insert into table2 values ('21:00:00.0000000');
insert into table2 values ('21:30:00.0000000');
以下SQL将为您提供免费插槽和学生姓名：
<强>查询：
select t1.StudentName,t2.timeslot
from Table2 t2,
Table1 t1
where t2.timeslot<t1.start_time
and t2.timeslot<t1.end_time
and t1.section='A'
group by t1.StudentName,t2.timeslot
order by t2.timeslot
<强>输出：
StudentName timeslot
1 Mary 09:00:00
2 Robert 09:00:00
3 Mary 09:30:00
4 Robert 09:30:00
5 Mary 10:00:00
6 Robert 10:00:00
7 Mary 10:30:00
8 Robert 10:30:00
9 Mary 11:00:00
10 Robert 11:00:00
11 Mary 11:30:00
12 Robert 11:30:00
13 Mary 12:00:00
14 Robert 12:00:00
15 Mary 12:30:00
16 Robert 12:30:00
17 Mary 13:00:00
18 Robert 13:00:00
这只是完成的一半任务，我只是告诉你实现它的方法。为了达到这个目的，在学生和科目表中再引入两个联接。
答案 2 :(得分：0)
将当天（09:00至21:30间隔）分成几分钟，找到关于该组学生的免费分钟数以及感兴趣的天数和分组间隔时间。
CREATE TABLE Table1 (ID varchar(10),StudentName varchar(30), Course varchar(15) ,SECTION varchar(2),DAY varchar(10),
START_TIME time , END_TIME time, actual_starttime time, actual_endtime time);
INSERT INTO Table1 VALUES (111, 'Mary','Science','A','Mon','13:30:00.0000000','16:20:00.0000000','09:00:00.0000000','21:20:00.0000000')
INSERT INTO Table1 VALUES (111, 'Mary','Maths','A','Tue','12:30:00.0000000','13:20:00.0000000','09:00:00.0000000','21:20:00.0000000')
INSERT INTO Table1 VALUES (111, 'Mary','Physics','C','Tue','10:30:00.0000000','11:10:00.0000000','09:00:00.0000000','21:20:00.0000000')
INSERT INTO Table1 VALUES (112, 'Robert','Maths','A','Mon','13:30:00.0000000','16:20:00.0000000','09:00:00.0000000','21:20:00.0000000')
;
-- parameters
declare @tds time = '09:00';
declare @tde time = '21:30';
declare @section varchar(2) = 'A';
create table #daysofinterest (DAY varchar(10) primary key);
insert #daysofinterest (DAY) values ('Mon'),('Tue'),('Fri');
create table #groupmembers(ID int primary key);
insert #groupmembers(ID) values (111),(112);
-- query
select DAY, startt = dateadd(minute, min(n), @tds), endt = dateadd (minute, max(n), @tds)
from (
select DAY, n, grp = n - row_number() over(partition by DAY order by n)
from (
-- all minutes of the day, @tds till @tde
select top (datediff(minute, @tds, @tde)) n = row_number() over(order by (select null))
from sys.all_objects
) tally
cross join #daysofinterest dd
join #groupmembers gm on
not exists (select 1 from table1 t
where t.ID = gm.ID and t.DAY = dd.DAY and SECTION = @section and
dateadd (minute, n, @tds) between t.START_TIME and t.END_TIME )
group by DAY, n
--this minute is free for every group member
having count(*) = (select count(*) from #groupmembers)
) g
group by DAY, grp
order by DAY, min(n)
答案 3 :(得分：0)
要返回可用时间段的完整列表，您需要为一周中的每一天构建一组所有时间段，然后查找这些时段中是否有任何一个在其中教授学生。
这可以通过递归CTE轻松实现，以构建完整的时间段集，您可以从中加入学生数据。以下查询的输出是每个空缺会议的日期和时间：
-- Build the dummy data sets:
declare @Data table
(
ID varchar(10),
StudentName varchar(30),
Course varchar(15),
SECTION varchar(2),
DAY varchar(10),
START_TIME time,
END_TIME time,
actual_starttime time,
actual_endtime time
);
insert into @Data values
(111, 'Mary', 'Science', 'A', 'Mon', '13:30:00.0000000', '16:20:00.0000000', '09:00:00.0000000', '21:20:00.0000000')
,(111, 'Mary', 'Maths', 'A', 'Tue', '12:30:00.0000000', '13:20:00.0000000', '09:00:00.0000000', '21:20:00.0000000')
,(111, 'Mary', 'Physics', 'C', 'Tue', '10:30:00.0000000', '11:10:00.0000000', '09:00:00.0000000', '21:20:00.0000000')
,(112, 'Robert', 'Maths', 'A', 'Mon', '13:30:00.0000000', '16:20:00.0000000', '09:00:00.0000000', '21:20:00.0000000');
-- Query the data:
with TimeSlots as -- Recursive CTE builds a table of all timeslots in TIME data type.
(
select cast('09:00:00' as time) as TimeSlotStart
,cast('09:30:00' as time) as TimeSlotEnd
union all
select dateadd(minute,30,TimeSlotStart)
,dateadd(minute,30,TimeSlotEnd)
from TimeSlots
where TimeSlotStart < cast('21:00:00' as time)
)
, TeachingDays as -- Used to return all the time slots above for each day of the week in CROSS JOIN below.
(
select 1 as DaySort
,'Mon' as TeachingDay
union all
select 2 as DaySort
,'Tue'
union all
select 3 as DaySort
,'Wed'
union all
select 4 as DaySort
,'Thu'
union all
select 5 as DaySort
,'Fri'
)
select td.TeachingDay
,t.TimeSlotStart
,t.TimeSlotEnd
from TimeSlots t -- Select all timeslots.
cross join TeachingDays td -- For each day.
left join @Data d -- And find all students that are being taught on that day at the specified time.
on(td.TeachingDay = d.DAY
and t.TimeSlotStart <= d.END_TIME
and t.TimeSlotEnd > d.START_TIME
)
where d.ID is null -- Then only return data where there are no students being taught at this timeslot.
order by td.DaySort
,t.TimeSlotStart;