我有Diary表的这个表结构:
CREATE TABLE Diary
(
[IdDiary] bigint,
[UserId] int,
[IdDay] numeric(18,0),
[IsAnExtraHour] bit
);
INSERT INTO Diary ([IdDiary], [UserId], [IdDay], [IsAnExtraHour])
values
(51, 1409, 1, 0),
(52, 1409, 1, 1),
(53, 1409, 3, 0),
(54, 1409, 5, 0),
(55, 1409, 5, 1),
(56, 1408, 2, 0);
这个结构为DiaryTimetable表:
CREATE TABLE DiaryTimetable
(
[IdDiary] bigint,
[Hour] varchar(50)
);
INSERT INTO DiaryTimetable ([IdDiary], [Hour])
VALUES
(51, '09:00'),
(51, '09:30'),
(51, '10:00'),
(51, '10:30'),
(51, '11:00'),
(52, '15:00'),
(52, '15:30'),
(52, '16:00'),
(52, '16:30'),
(52, '17:00'),
(53, '11:00'),
(53, '11:30'),
(53, '12:00'),
(53, '12:30'),
(53, '13:00'),
(54, '10:00'),
(54, '10:30'),
(54, '11:00'),
(54, '11:30'),
(54, '12:00'),
(55, '16:00'),
(55, '16:30'),
(55, '17:00'),
(55, '17:30'),
(55, '18:00'),
(56, '15:00'),
(56, '15:30'),
(56, '16:00'),
(56, '16:30'),
(56, '17:00');
我使用此查询获取用户ID 1409的最大小时和最小时数,以获取每天输入的时间和离开工作的时间。 idday与星期几的数字相对应。例如1是星期一,2是星期二等...
SELECT d.IdDiary, d.IdDay, MIN(Hour) as 'Start Time', MAX(Hour) as 'End Time', IsAnExtraHour
FROM Diary AS d
LEFT JOIN DiaryTimetable AS dt ON d.IdDiary = dt.IdDiary
where userid = 1409
GROUP BY d.IdDiary, d.IdDay, IsAnExtraHour
此查询提供此结果:
我想得到这个结果:
Day Start Time End Time Start Extra Time End Extra Time
----- ---------- -------- --------------- ---------------
Monday 09:00 11:00 15:00 17:00
Wednessday 11:00 13:00
Friday 10:00 12:00 16:00 18:00
我有一个列(IsAnExtraHour)此列表示此行是否在一天中有额外的小时数,例如员工在星期一09:00到11:00开始工作,然后在下午15:00再次工作到了17:00,所以我想知道如何将这个小时分组在同一行,我希望我能够清楚地表达,我接受建议表示感谢。
答案 0 :(得分:9)
SELECT d.IdDay,
MIN(CASE WHEN isAnExtraHour = 0 THEN hour END) as 'Start Time',
MAX(CASE WHEN isAnExtraHour = 0 THEN hour END) as 'End Time',
MIN(CASE WHEN isAnExtraHour = 1 THEN hour END) as 'Start Extra Time',
MAX(CASE WHEN isAnExtraHour = 1 THEN hour END) as 'End Extra Time'
FROM Diary AS d
LEFT JOIN
DiaryTimetable AS dt
ON dt.IdDiary = d.IdDiary
WHERE userid = 1409
GROUP BY
d.IdDay
答案 1 :(得分:5)
我使用了来自@Quassnoi的代码,我添加了这个:
SELECT DATENAME(weekday, d.idday-1) as 'Day' ,
MIN(CASE WHEN isAnExtraHour = 0 THEN hour END) AS 'Start Time',
MAX(CASE WHEN isAnExtraHour = 0 THEN hour END) AS 'End Time',
MIN(CASE WHEN isAnExtraHour = 1 THEN hour END) AS 'Start Extra Time',
MAX(CASE WHEN isAnExtraHour = 1 THEN hour END) AS 'End Extra Time'
FROM Diary AS d
LEFT JOIN DiaryTimetable AS dt ON dt.IdDiary = d.IdDiary
WHERE userid = 1409
GROUP BY d.IdDay
我希望这能帮到别人,谢谢大家的回答。
答案 2 :(得分:1)
我接近这个的方法是分别计算你的标准和额外时间,就像这样;
SELECT d.IdDay
,MIN(dt.StartTime) AS 'Start Time'
,MAX(dt.EndTime) AS 'End Time'
,MIN(ex.StartTime) AS 'Start Extra Time'
,MAX(ex.EndTime) AS 'End Extra Time'
FROM Diary AS d
LEFT JOIN (
SELECT IdDiary
,MIN(Hour) AS StartTime
,MAX(Hour) AS EndTime
FROM DiaryTimetable
GROUP BY IdDiary
) AS dt ON d.IdDiary = dt.IdDiary
AND d.IsAnExtraHour = 0
LEFT JOIN (
SELECT IdDiary
,MIN(Hour) AS StartTime
,MAX(Hour) AS EndTime
FROM DiaryTimetable
GROUP BY IdDiary
) AS ex ON d.IdDiary = ex.IdDiary
AND d.IsAnExtraHour = 1
WHERE userid = 1409
GROUP BY d.IdDay
答案 3 :(得分:1)
您可以先获得非额外时间,然后通过UserId和IdDay将加入时间加入他们。像这样:
SELECT
CASE BaseHours.IdDay
WHEN 1 THEN 'Monday'
WHEN 2 THEN 'Tuesday'
WHEN 3 THEN 'Wednesday'
WHEN 4 THEN 'Thursday'
WHEN 5 THEN 'Friday'
WHEN 6 THEN 'Saturday'
WHEN 7 THEN 'Sunday'
END AS [WeekDay],
MIN(BaseTimeTable.Hour) as 'Start Time',
MAX(BaseTimeTable.Hour) as 'End Time',
MIN(ExtraTimeTable.Hour) as 'Start Extra Time',
MAX(ExtraTimeTable.Hour) as 'End Extra Time'
FROM Diary AS BaseHours
LEFT JOIN Diary ExtraHours
ON ExtraHours.UserId = BaseHours.UserId
AND ExtraHours.IdDay = BaseHours.IdDay AND ExtraHours.IsAnExtraHour = 1
JOIN DiaryTimetable AS BaseTimeTable
ON BaseHours.IdDiary = BaseTimeTable.IdDiary
LEFT JOIN DiaryTimetable AS ExtraTimeTable
ON ExtraHours.IdDiary = ExtraTimeTable.IdDiary
WHERE BaseHours.Userid = 1409 AND BaseHours.IsAnExtraHour = 0
GROUP BY BaseHours.IdDay