Question

问题：对于特定患者，在接下来的7天内不会对呼叫进行后续预约

我的查询：

  select *, DATEDIFF(DAY, (APPOINTMENT_DATE - LAG(APPOINTMENT_DATE) 
  over (ORDER BY PATIENT_ID)), APPOINTMENT_DATE) as DIFFERENCE from [dbo]. 
  [Appointment Data]

问题：

1.由于日期时间的原因，某些疯狂格式的差异可能会改变。

2.我的查询对吗？如何找到每个客户的差异？我知道我必须申请分组，但我有点困惑。

PLS帮助！

数据集：

APPOINTMENT_DATE             PATIENT_ID         DIFFERENCE                                                  

2010-05-06 00:00:00.000      00051101          NULL
2010-04-11 00:00:00.000      00101005          40302
2010-05-06 00:00:00.000      00130521          40277
2010-02-07 00:00:00.000      00130521          40302

Answer 1

select *,
DATEDIFF(DAY, LAG(APPOINTMENT_DATE) over (ORDER BY PATIENT_ID), APPOINTMENT_DATE) as DIFFERENCE
from [dbo].[Appointment Data]

结果：

+-----------------------+------------+------------+
|   APPOINTMENT_DATE    | PATIENT_ID | DIFFERENCE |
+-----------------------+------------+------------+
| 5/6/2010 12:00:00 AM  |   00051101 |       null |
| 4/11/2010 12:00:00 AM |   00101005 |        -25 |
| 5/6/2010 12:00:00 AM  |   00130521 |         25 |
| 2/7/2010 12:00:00 AM  |   00130521 |        -88 |
+-----------------------+------------+------------+

如果您切换日期，结果会有所不同。

select *,
DATEDIFF(DAY, APPOINTMENT_DATE, LAG(APPOINTMENT_DATE) over (ORDER BY PATIENT_ID)) as DIFFERENCE
from [dbo].[Appointment Data]

结果：

+-----------------------+------------+------------+
|   APPOINTMENT_DATE    | PATIENT_ID | DIFFERENCE |
+-----------------------+------------+------------+
| 5/6/2010 12:00:00 AM  |   00051101 |       null |
| 4/11/2010 12:00:00 AM |   00101005 |         25 |
| 5/6/2010 12:00:00 AM  |   00130521 |        -25 |
| 2/7/2010 12:00:00 AM  |   00130521 |         88 |
+-----------------------+------------+------------+

Answer 2

您的查询中似乎有多处错误：

1）您应该在分区中使用列PATIENT_ID，并在APPOINTMENT_DATE函数中按LAG排序

2）您在DATEDIFF函数

中有不必要的减法

因此，您的查询应该是：

select 
    *, datediff(dd, lag(APPOINTMENT_DATE) over (partition by PATIENT_ID order by APPOINTMENT_DATE), APPOINTMENT_DATE)
from 
    [dbo].[Appointment Data]

日期差异在SQL Server中使用LAG函数时更改为42000

2 个答案: