识别具有相同 ID 的列

Identify the columns that have the same ID

我想创建一个新列,在本例中为“Reiterate”,它为具有相同键 (Clave) 并且在 resolutiondate 和 [= 之间的时间少于 7 天的行标记为“是” 12=],并且 Massive 列必须为“NO”。这是我想要得到的例子。

我想尝试使用 lead and lag 公式将日期从最低到最高排序,但不知道如何在 SQL 中应用它。我不太确定是否有可能获得该专栏。

非常感谢您的帮助。非常感谢

编辑。

我使用了您建议的公式并且有效! , 再次感谢所有帮助:)

SELECT top 100
CF1.STRINGVALUE AS 'ID OT',
jis.created, 
jis.resolutiondate,
IIF(CF3.STRINGVALUE like 'IDR-%','SI','NO') AS 'Massive',
cfo8.customvalue AS 'Solved',
CASE WHEN jis.resolutiondate IS NULL 
     THEN 'NO'
     WHEN LAG(resolutiondate) OVER (PARTITION BY 'ID OT' ORDER BY resolutiondate) >  DATEADD(DAY, -7, resolutiondate)
     AND LAG(CF3.STRINGVALUE) OVER (PARTITION BY 'ID OT' ORDER BY resolutiondate) = 'NO'
     AND LAG(cfo8.customvalue) OVER (PARTITION BY 'ID OT' ORDER BY resolutiondate) = 'NO'
     THEN 'NO'
     ELSE 'YES'
END AS SLA

FROM [DWH].[JIR].[jiraissue] jis
LEFT JOIN [DWH].[JIR].[customfieldvalue] CF1 ON (CF1.issue = jis.id AND CF1.CUSTOMFIELD = 10004)
LEFT JOIN [DWH].[JIR].[customfieldvalue] CF3 ON (CF3.issue = jis.id AND CF3.CUSTOMFIELD = 10032)
LEFT JOIN [DWH].[JIR].[customfieldvalue] CF14 ON (CF14.issue = jis.id AND CF14.CUSTOMFIELD = 10906)
LEFT JOIN [DWH].[JIR].customfieldoption cfo8 ON (CF14.customfield = cfo8.customfield AND CF14.stringvalue=CAST(cfo8.id AS CHAR))

如果我没听错,你可以使用window功能:

select t.*,
    case 
        when lag(resolutiondate) over(partition by key order by resolutiondate) > dateadd(day, -7, resolutiondate)
            and lag(massive) over(partition by key order by resolutiondate) = 'NO'
        then 'SI'
        else 'NO'
    end as sla
from mytable t

这会将当前行的日期与前一行的日期进行比较,并检查前一行的 massive 状态。

也许您想检查当前行的状态而不是上一行的状态 - 这有点简单:

select t.*,
    case 
        when lag(resolutiondate) over(partition by key order by resolutiondate) > dateadd(day, -7, resolutiondate)
            and massive = 'NO'
        then 'SI'
        else 'NO'
    end as sla
from mytable t

我想这就是你想要的逻辑:

select t.*,
       (case when massive = 'SI' then 'NO'
             when lag(resolutiondate) over (partition by key) < 
                  dateadd(-7, day, creationdate
             then 'No'
             else 'Si'
        end) as SLA
from t;

按照这种写法,似乎没有必要计算键列。每个密钥的解决日期都相同。像这样

select t.*,
       case when t.massive='No' and dt.diff<7
            then 'Si'
            else 'No' end SLA
from tTable t
     cross apply (values (datediff(day, created, resolutiondate))) dt(diff);