问题描述
我希望计算每个人的连续日法术.
I am looking to count consecutive day spells for each individual person.
我的桌子:
CREATE TABLE Absence( Date Date, Code varchar(10), Name varchar(10), Type varchar(10) ); INSERT INTO Absence (Date, Code, Name, Type) VALUES ('01-10-18', 'S', 'Sam', 'Sick'), ('01-11-18','S', 'Sam', 'Sick'), ('01-12-18','S', 'Sam', 'Sick'), ('01-21-18','S', 'Sam', 'Sick'), ('01-26-18','S', 'Sam', 'Sick'), ('01-27-18','S', 'Sam', 'Sick'), ('02-12-18','S', 'Sam', 'Holiday'), ('02-13-18','S', 'Sam', 'Holiday'), ('02-18-18','S', 'Sam', 'Holiday'), ('02-25-18','S', 'Sam', 'Holiday'), ('02-10-18','S', 'Sam', 'Holiday'), ('02-13-18','F', 'Fred', 'Sick'), ('02-14-18','F', 'Fred', 'Sick'), ('02-17-18','F', 'Fred', 'Sick'), ('02-25-18','F', 'Fred', 'Sick'), ('02-28-18','F', 'Fred', 'Sick');
这是我目前拥有的代码:
This is the code i currently have:
WITH CTE AS ( SELECT Date, Name, Type ,GroupingSet = DATEADD(DAY, ROW_NUMBER() OVER (PARTITION BY [Name], [Type] ORDER BY [Date]), [Date]) FROM Absence ) SELECT Name, StartDate = MIN(Date), EndDate = MAX(Date), Result = COUNT(Name), min(Type) AS [Type] FROM CTE GROUP BY Name, GroupingSet -- HAVING COUNT(NULLIF(Code, 0)) > 1 ORDER BY Name, StartDate
产生的结果:
| Name | StartDate | EndDate | Result | Type | |------|------------|------------|--------|---------| | Fred | 2018-02-13 | 2018-02-13 | 1 | Sick | | Fred | 2018-02-14 | 2018-02-14 | 1 | Sick | | Fred | 2018-02-17 | 2018-02-17 | 1 | Sick | | Fred | 2018-02-25 | 2018-02-25 | 1 | Sick | | Fred | 2018-02-26 | 2018-02-28 | 1 | Sick | | Sam | 2018-01-10 | 2018-01-10 | 1 | Sick | | Sam | 2018-01-11 | 2018-01-11 | 1 | Sick | | Sam | 2018-01-12 | 2018-01-12 | 1 | Sick | | Sam | 2018-01-21 | 2018-01-21 | 1 | Sick | | Sam | 2018-01-26 | 2018-01-26 | 1 | Sick | | Sam | 2018-01-27 | 2018-01-27 | 1 | Sick | | Sam | 2018-02-10 | 2018-02-10 | 1 | Holiday | | Sam | 2018-02-12 | 2018-02-12 | 1 | Holiday | | Sam | 2018-02-13 | 2018-02-13 | 1 | Holiday | | Sam | 2018-02-18 | 2018-02-18 | 1 | Holiday | | Sam | 2018-02-25 | 2018-02-25 | 1 | Holiday |
我正在寻找这样的结果集:
Where as i am looking for a result set like this:
| Name | Date | Result | Type | |------|------------|---------|---------| | Fred | 2018-02-13 | 2 | Sick | | Sam | 2018-01-27 | 2 | Sick | | Sam | 2018-02-10 | 1 | Holiday |
我需要计算连续超过 1 天的连续天数.然后把这个作为一个人有多少连续法术的总数.例如在那段时间里,弗雷德连续两次生病.如果有人周五和周一休息,我也需要这个来覆盖,这应该算作连续的咒语.
I need to count the consecutive days where there is more then 1 day in a row. And then have this as a total of how many consecutive spells someone had. e.g. fred had 2 consecutive sick spells during that time period. I also need this to cover if someone had a friday and a monday off, this should count as a consecutive spell.
我对如何到达那里有点迷茫.任何帮助将不胜感激.
Im a bit lost as to how to get there. Any help would be appreciate.
请参阅:http://sqlfiddle.com/#!18/88612/16
推荐答案
您可以使用以下方法获取缺勤时间:
You can get the periods of absences using:
select name, min(date), max(date), count(*) as numdays, type from (select a.*, row_number() over (partition by name, type order by date) as seqnum_ct from absence a ) a group by name, type, dateadd(day, -seqnum_ct, date);
这里是一个 SQL Fiddle.
Here is a SQL Fiddle for this.
你可以添加有count(*)>1 获得一天或更长时间的经期.这似乎很有用.我不明白最终的输出是什么.描述对我来说没有意义.
You can add having count(*) > 1 to get periods with one day or more. This seems useful. I don't understand what the ultimate output is. The description just doesn't make sense to me.
如果您想要 2 天或更多天的缺勤次数,请将其用作子查询/CTE:
If you want the number of absences that are 2 or more days, then use this as a subquery/CTE:
select name, count(*), type from (select name, min(date) as mindate, max(date) as maxdate, count(*) as numdays, type from (select a.*, row_number() over (partition by name, type order by date) as seqnum_ct from absence a ) a group by name, type, dateadd(day, -seqnum_ct, date) ) b where numdays > 1 group by name, type;