Question

我写了一个查询，返回一组代理的开放支持票数。查询当前返回的时间超过5天：

SELECT groupname, 
       assignee, 
       Count(*) AS [Tickets Older Than 5 Days] 
FROM   zendeskunclosedticketexport 
WHERE  ( groupname = 'DC Web Ops' 
          OR groupname = 'DC Maintenance' 
          OR groupname = 'DC QC1' ) 
       AND ticketcreateddate < Dateadd(day, -5, Getdate()) 
GROUP  BY groupname, 
          assignee 
ORDER  BY groupname, 
          Count(*) DESC

查询结果示例：

groupname       assignee                Tickets Older Than 5 Days

DC Maintenance  Mubarak Ali             12
DC Maintenance  Abirami Selvanathan     11
DC Maintenance  Jemima Mohanraj         6
DC Maintenance  Praveena Rajan          6
DC Maintenance  Rahul Ramesh            6
DC Maintenance  Saravanan Varadharajan  6
DC Web Ops  Kasey Barber                16
DC Web Ops  Jamie Raines                11
DC Web Ops  Keith Washington            10
DC Web Ops  Patricia Gomez Imvendorize  6

我一直试图做的还包括10天，30天等的列，但一直无法做到。我试图在select语句中编写一个子查询来返回相应天数的计数;如下所示：

SELECT groupname, 
       assignee [currentAssignee],


       (SELECT Count(*) AS [Tickets Older Than 5 Days] 
            FROM   zendeskunclosedticketexport 
            WHERE  ( 
                    ticketcreateddate < Dateadd(day, -5, Getdate())
                    AND
                    assignee = [currentAssignee]
        )

       (SELECT Count(*) AS [Tickets Older Than 10 Days] 
            FROM   zendeskunclosedticketexport 
            WHERE  ( 
                    ticketcreateddate < Dateadd(day, -10, Getdate())
                    AND
                    assignee = [currentAssignee]
        )

        (SELECT Count(*) AS [Tickets Older Than 30 Days] 
            FROM   zendeskunclosedticketexport 
            WHERE  ( 
                    ticketcreateddate < Dateadd(day, -30, Getdate())
                    AND
                    assignee = [currentAssignee]
        )
FROM   zendeskunclosedticketexport

我在这里尝试做的是将外部查询中的受理人值传递给内部查询...我知道，因为我拥有它这不合法，但希望它提供了一些关于方法的背景我在拿。有没有办法以临时变量的形式将数据从外部查询传递到内部查询？

编辑：用户通知我，我错过了一个FROM子句，之后添加了这个。

Answer 1

您可以使用SUM CASE语句更简单地执行此操作：

SELECT
    groupname,
    assignee,
    SUM(CASE WHEN ticketcreateddate < DATEADD(DAY, -5, GETDATE()) THEN 1 ELSE 0 END) AS [Tickets Older Than 5 Days],
    SUM(CASE WHEN ticketcreateddate < DATEADD(DAY, -10, GETDATE()) THEN 1 ELSE 0 END) AS [Tickets Older Than 10 Days],
    SUM(CASE WHEN ticketcreateddate < DATEADD(DAY, -30, GETDATE()) THEN 1 ELSE 0 END) AS [Tickets Older Than 30 Days]
FROM
    zendeskunclosedticketexport
WHERE
    groupname IN ('DC Web Ops', 'DC Maintenance', 'DC QC1')
GROUP BY
    groupname,
    assignee
ORDER BY
    groupname

您可以将ORDER BY设置为您需要的那些聚合列中的任何一个。

Select语句中的子查询

1 个答案: