识别具有相同 ID 的列

Posted 2023-05-08

【中文标题】识别具有相同 ID 的列

【英文标题】：Identify the columns that have the same ID

【发布时间】：2020-12-24 11:57:30

【问题描述】：

我想创建一个新列，在这种情况下，“Reiterate”为具有相同键 (Clave) 并且在 resolutiondate 和 createddate 之间的时间少于 7 天的行标记为“YES”，并且Massive 列必须为“否”。这是我想要得到的示例。

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。我不确定是否有可能获得该专栏。

非常感谢任何帮助。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))

【参考方案1】：

我认为这是你想要的逻辑：

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;

【参考方案2】：

如果我没听错，你可以使用窗口函数：

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上一行的日期进行比较，并检查上一行的massive 状态。

也许您想检查当前行而不是前一行的状态 - 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

【参考方案3】：

按照这种编写方式，似乎没有必要评估键列。每个密钥的解决日期相同。像这样的

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);