Question

首先在这里发帖，但任何人都可以指导或帮助我理解以下问题。对于下表＆＃34; Patient_Table＆＃34;，我怎样才能找出patient_id 22生病的总天数。

ID    Patient_ID     Listing_Number     Date              Status 
-----------------------------------------------------------------
1     22              1                  01/01/2016        Healthy
2     22              2                  01/11/2016        Sick
3     34              1                  01/13/2016        Healthy
4     22              3                  01/20/2016        Healthy
5     22              4                  01/22/2016        Sick
6     22              5                  01/23/2016        Healthy

到目前为止，

是我的逻辑，但我不确定正确的语法。

declare 
@count      int = 1, 
@Days_sicks int = 0 

while   @count <= (select max (Listing_number) from Patient_Table where Patient_id = '22')

begin 
    case
    when (select status from Patient_Table where Patient_id = '22' and Listing_number = @count) = 'Sick'
    then @Days_sicks = @Days_sicks + datediff('dd',  (select min date from Patient_Table where patient_id = 22 and listing_number >  @count and status != 'Sick'), (select date from patient_table where patient_id = 22 and listing_number = @count)
    else   @Days_sicks 
    end as Days_sicks

set @Count = @Count + 1 
END;

我也试过这个，但是效果不好而且我对这个小组的条款有问题

SELECT t1.patient_id, 
    DATEDIFF(dd,
        t1.date,
        (SELECT MIN(t3.date)
        FROM Patient_Table t3
        WHERE t3.patient_id = t1.patient_id
        AND t3.date> t1.date) as Days_sicks
    )
FROM Patient_Table t1
WHERE EXISTS(
    SELECT 'NEXT'
    FROM Patient_Table t2
   WHERE t2.patient_id = t1.patient_id
   AND t2.date> t1.date
   AND t2.status != 'sick')
   and t1.patient_id = '22'

期望的结果

Patient id    Days_sicks
22            10

Answer 1

使用lead()功能，然后聚合：

select patient_id,
       sum(datediff(day, date, coalesce(next_date, getdate())))
from (select pt.*,
             lead(date) over (partition by patient_id order by date) as next_date
      from patient_table pt
     ) pt
where status = 'Sick'
group by patient_id;

注意：如果有人当前生病了，那么这会使用当前日期结束＆＃34;生病＆＃34;状态。

此外，如果患者多次就诊，这将起作用。

Answer 2

Select s.Patient_ID, 
   sum(datediff(day, s.Date, coalesce(h.date, getdate()) totalDaysSick 
From table s left join table h 
   on h.Patient_ID = s.Patient_ID 
     and h.Status = 'Healthy'
     and s.Status ='Sick'  
     and h.Date = (Select Min(date) from table 
                   where Patient_ID = s.Patient_ID
                     and Date > s.Date)
Group By Patient_Id

Answer 3

您可以使用Lag窗口功能并按日期排序，然后执行日期更改以查找当前日期和上一日期之间的天数。然后对天数求一笔：

select
    sickdays.Patient_ID
    ,Sum(sickdays.DayNums) as SickDays
from
    (select
            Patient_ID
            ,Datediff(day, Lag([Date], 1, null) over (order by [Date]), [Date]) as DayNums

        from Patient
        where Patient_ID = 22
        and Status = 'Sick'
        group by    Patient_ID
                    ,Date) as sickdays
where
    sickdays.DayNums is not null
group by sickdays.Patient_ID

Microsoft SQL Server计算连续事件之间的总时间

3 个答案: