Question

我添加了一张我遇到的照片。突出显示的是我需要抓住的，带删除线的那些是我不需要的。所以第一个胆小者（561346）没有6个月的间隔，所以我希望所有这些行。第二个Tempid（661450）在第9行有6个月的间隔，所以我只想要下一个（第10行）该不需要的那个剩余的那行。 Last tempid（662020）在第16行有8个月的间隙，然后在其后还有4行。我只想抓住那4行。我希望这是有道理的。

我只需要一个人来帮助我。我已经尝试了好几天了。

  WITH cte AS 
  ( 
  SELECT pt.TempID, MIN(pt.InvoiceDate) AS 'min', MAX(pt.InvoiceDate) AS 
  'Max', 
  pt.JobNumber , 
  ROW_NUMBER() OVER (PARTITION BY tempid ORDER BY MAX(pt.InvoiceDate)) AS 
  'Row'  
  FROM #temp t 
  JOIN pamwebhouse..pamtempjobs pt 
  ON SUBSTRING(t.EMPLOYEE, 5,10) = pt.TempID 
  WHERE 
  pt.Status <> 'Posted'
  AND pt.PayTypeID IN ('Regul','OT','Sick') 
  GROUP BY pt.TempID, pt.jobnumber
  ) 
  SELECT a.TempID, a.JobNumber,DATEDIFF(month, a.max, a.start) AS 'Diff' , 
  A.MAX  
  , a.Row  
  INTO #temp2 
  FROM ( 
  SELECT c.TempID, c.JobNumber, c.min AS 'Min', c.max AS 'Max', 
  CASE WHEN c2.TempID IS NULL THEN NULL ELSE c2.min END AS 'Start', c.Row
  FROM cte c
   LEFT JOIN cte c2 
   ON c2.row = c.Row +1
   AND c.TempID = c2.TempID
   --ORDER BY c.TempID
   ) a 

   ORDER BY a.TempID


    --This is what is giving me the output in the screenshot. 
    select t.tempid, t.JobNumber, test AS 'Gap',row,
             max(case when test >= 6 then jobnumber  END) over (PARTITION BY 
    tempid ORDER by max) as id_6
      from #temp2 t

documentation

Answer 1

您可以使用窗口功能：

select t.*
from (select t.*,
             max(case when gap = 6 then row end) over (partition by tempid order by row) as seq
      from table t
     ) t
where (t.gap <> 6 and t.seq is not null);

找到6个月的时间间隔并抓住所有东西之后，如果没有6个月的时间间隔，那么就抓住所有东西

1 个答案: