Question

我觉得这应该是一件容易的事，但我正在努力。

我有一个问题表，表中有以下内容：

用户ID
创建日期
已解决的日期
发行编号

我通过用户ID总结了这一点，其中包含日期创建的问题数和日期解决的问题数。

例如，

User ID   ///   Date     ///     Issues Received  ///  Issues Resolved
JonSmith1 ///  1/12/2015 ///            28        ///          34

问题源于这样一个事实，即在我计算已解决的问题时，我并不关心何时收到问题。因此，昨天，上周，去年等都可以收到已解决计算中的问题。

非常感谢任何帮助。

谢谢， PE

Answer 1

对这两个日期执行单独查询，然后将它们与UNION：

合并

SELECT UserID, Date, MAX(IssuesReceived) AS IssuesReceived, MAX(IssuesResolved) AS IssuesResolved
FROM (
    SELECT UserID, CreatedDate AS Date, COUNT(*) AS IssuesReceived, 0 AS IssuesResolved
    FROM Issues
    GROUP BY UserID, Date
    UNION
    SELECT User, ResolvedDate AS Date, 0 AS IssuesReceived, COUNT(*) AS IssuesResolved
    FROM Issues
    GROUP BY UserID, Date) AS u
GROUP BY UserID

Answer 2

试试这个。

select userid,createddate as date,
      count(`created date`) as `Issues Received`,
      count(`Resolved Date`) as `Issues Resolved`
from yourtable 
where `Resolved Date` = '1/12/2015'
Group by userid,createddate

考虑到创建新问题时解决的日期将为空

Answer 3

不确定这是否完全回答了你的问题，但很可能你需要做这样的联盟：

select user_id
    , action_date
    , count(case when action = 'CREATED' then 1 else null end)
    , count(case when action = 'RESOLVED' then 1 else null end)
from (
     select user_id, created_date action_date, 'CREATED' action
     from table
     union
     select user_id, resolved_date, 'RESOLVED' action
     from table
     where resolved_date is not null
     ) x
group by user_id, action_date

这有用吗？

Answer 4

如果您确实使用的是Teradata，则可以使用full outer join：

SELECT COALESCE(i.UserID, r.UserId) as UserId,
       COALESCE(i.Date, r.Date) as Date,
       COALESCE(IssuesReceived, 0) AS IssuesReceived, 
       COALESCE(IssuesResolved, 0) AS IssuesResolved
FROM (SELECT UserID, CreatedDate AS Date, COUNT(*) AS IssuesReceived
      FROM Issues
      GROUP BY UserID, Date
     ) i FULL OUTER JOIN
     (SELECT User, ResolvedDate AS Date, COUNT(*) AS IssuesResolved
      FROM Issues
      GROUP BY UserID, Date
     ) r
     ON i.UserId = r.UserId and i.Date = r.Date;

需要从表中的两个不同日期开始计算

4 个答案: