Question

即使我找到了类似于我所需要的东西，但我还是无法解决这个问题。

所以，我会问我的例子。

（继续我的上一个问题：select if there are more than n results but with conditions）

我的表：


| TimeId | Work_Role | User_Name |
----------------------------------
|    1   | users     | Oran      | 
|    2   | admin     | Ray       |
|    3   | users     | Oran      | 
|    4   | servs     | Amit      |
|    5   | admin     | Oran      |
|    6   | users     | Ray       |
|    7   | users     | Oran      | 
|    8   | servs     | Amit      |
|    9   | admin     | Oran      |
|   10   | users     | Oran      |

我已经使用CTE显示了user_name =＆＃34; Oran＆＃34;的列表。和Work_Role =＆＃34;用户＆＃34;只要表格中有超过2个。

此外，我使用Union all来收集另一部分，其条件是：user_name =＆＃34; Oran＆＃34;和Work_Role =＆＃34; admin＆＃34;。

结果是：

| TimeId | Work_Role | User_Name |
----------------------------------
|    1   | users     | Oran      | 
|    3   | users     | Oran      | 
|    7   | users     | Oran      | 
|   10   | users     | Oran      |
|    5   | admin     | Oran      | 
|    9   | admin     | Oran      |

我的第二部分（我有问题）是计算上面结果有多少行。

在这种情况下，我希望：6

我的第一个结果的代码（工作正常）：

WITH CTE AS(
    SELECT *, cc = COUNT(*) OVER(PARTITION BY User_Name) FROM Table
)
SELECT *
FROM CTE
WHERE
    Work_Role = 'users' 
    AND User_Name = 'Oran'
    AND cc > 2
UNION ALL
SELECT * 
FROM TABLE
WHERE
   Work_Role = 'admin' 
   AND User_Name = 'Oran'

我如何总结这些行＆＃39;最终结果的数量？

使用CTE时是否可以完成？

注意： 这只是一个例子。在我的真实代码中，我必须使用＆＃34; UNION ALL＆＃34; ： - （

10X！

Answer 1

首先，您可以将查询简化为：

WITH CTE AS (
      SELECT t.*, cc = COUNT(*) OVER(PARTITION BY User_Name) FROM Table t
     )
SELECT *
FROM CTE
WHERE (Work_Role = 'users' AND User_Name = 'Oran' AND cc > 2) OR
      (Work_Role = 'admin' AND User_Name = 'Oran');

好处是它也可以工作，因为这两个子查询具有相同的列。

如果您想计算行数，请使用select count(*)：

WITH CTE AS (
      SELECT t.*, cc = COUNT(*) OVER(PARTITION BY User_Name) FROM Table t
     )
SELECT COUNT(*)
FROM CTE
WHERE (Work_Role = 'users' AND User_Name = 'Oran' AND cc > 2) OR
      (Work_Role = 'admin' AND User_Name = 'Oran');

Answer 2

好的，最终我找到了答案，哈哈，

这是基于这个问题的第一个答案：Sum a union query

我必须为我的派生表添加别名。
在“THIS”和“SELECT”之间设置附加选择（使用SUM）。
为每个查询更改为COUNT并为其添加别名。

像这样：

WITH CTE AS(
    SELECT *, cc = COUNT(*) OVER(PARTITION BY User_Name) FROM Table
)
SELECT SUM(my_dervied_table.total_lines) FROM (
SELECT COUNT(*) as 'total_lines'
FROM CTE
WHERE
    Work_Role = 'users' 
    AND User_Name = 'Oran'
    AND cc > 2
UNION ALL
SELECT  COUNT(*) as 'total_lines' 
FROM TABLE
WHERE
   Work_Role = 'admin' 
   AND User_Name = 'Oran'
) my_dervied_table

所以不要计算此输出中的行：

| TimeId | Work_Role | User_Name |
----------------------------------
|    1   | users     | Oran      | 
|    3   | users     | Oran      | 
|    7   | users     | Oran      | 
|   10   | users     | Oran      |
|    5   | admin     | Oran      | 
|    9   | admin     | Oran      |

我实际上在总结'total_lines'列中“my_derived_table”的内容：4 + 2 = 6

    | total_lines |
    ---------------
 1) |      4      |
 2) |      2      |

Answer 3

shalom yael-试试这段代码 -

   WITH CTE AS 

(
SELECT *
FROM tbl
WHERE
    Work_Role = 'users' 
    AND User_Name = 'Oran'
    AND ( select COUNT(user_name)  
          FROM tbl
          WHERE user_name='oran') > 2
UNION ALL
SELECT * 
FROM tbl
WHERE
   Work_Role = 'admin' 
   AND User_Name = 'Oran')
   SELECT count([User_Name]) total_lines  
   from cte
   GROUP BY [Work_Role],[User_Name]

Answer 4

你在找这个，

Declare @t table(TimeId int,Work_Role varchar(50),User_Names varchar(50))
insert into @t values
 (1  , 'users','Oran'      ), 
(   2 ,'admin', 'Ray'),
(   3 ,'users', 'Oran'),
(    4,'servs', 'Amit'),
(    5,'admin', 'Oran'),
(    6,'users', 'Ray'),
(    7,'users', 'Oran'),
(    8,'servs', 'Amit'),
(    9,'admin', 'Oran'),
(   10,'users', 'Oran')
;With CTE as
(
   select *,
   ROW_NUMBER()over(partition by User_Names,Work_Role order by timeid)rn 
   from @t
)
   select * from CTE where Work_Role = 'users' 
    AND User_Names = 'Oran'
    AND exists
    (
    select User_Names from CTE where rn>2 
    and Work_Role = 'users' AND User_Names = 'Oran'
    )

 UNION ALL
  select * from CTE where Work_Role = 'admin' 
  AND User_Names = 'Oran'

SUM查询的结果包含UNION ALL和CTE

4 个答案: