减去两个子查询答案

【问题标题】：Subtract two subqueries减去两个子查询
【发布时间】：2013-01-02 15:07:46
【问题描述】：

我有以下疑问。我想从 SUM 中减去数字，然后只返回结果不是 0 的数字。任何帮助将不胜感激。我也很想看看这是否可以在一个查询中完成。我用 HAVING 子句尝试了它，但它返回了错误的结果。使用 SQL Server 2008

  SELECT 
(
 SELECT station_id, SUM(tcl_missing + tcl_not_missing) as tcl_total
 FROM tcl_missing_summary 
 GROUP BY station_id
) as a 

(
 SELECT station_id, SUM(total) as total
 FROM tcl_breakdown_op
 WHERE tr_standard not like '%cru'
 GROUP BY station_id
) as b
 WHERE a.tcl_total - total <> 0

【问题讨论】：

标签： sql sql-server select subquery

【解决方案1】：

假设您有多个 stationID，

  SELECT station_id, tcl_total, total, a.tcl_total - total as diff
from (
 SELECT station_id, SUM(tcl_missing + tcl_not_missing) AS tcl_total
 FROM tcl_missing_summary 
 GROUP BY station_id
) AS a INNER JOIN

(
 SELECT station_id, SUM(total) AS total
 FROM tcl_breakdown_op
 WHERE tr_standard NOT LIKE '%cru'
 GROUP BY station_id
) AS b
ON a.stationid = b.stationid
 WHERE a.tcl_total - total <> 0

也可以考虑

SELECT 
    station_id,
    SUM(tcl_missing + tcl_not_missing) AS sum_tcl_total,
    SUM(total) AS sum_total,
    SUM(tcl_missing + tcl_not_missing) - SUM(total) AS diff
FROM 
    tcl_missing_summary s INNER JOIN
    tcl_breakdown_op b ON
    s.station_id = b.station_id
WHERE
     tr_standard NOT LIKE '%cru'    
GROUP BY 
    station_id
HAVING
    SUM(tcl_missing + tcl_not_missing) <> SUM(total)

【讨论】：

【解决方案2】：

试试这个

; WITH CTE1 AS 
(
SELECT station_id, SUM(tcl_missing + tcl_not_missing) as tcl_total
 FROM tcl_missing_summary 
 GROUP BY station_id
), CTE2 AS
(
SELECT station_id, SUM(total) as total
 FROM tcl_breakdown_op
 WHERE tr_standard not like '%cru'
 GROUP BY station_id
)

SELECT * FROM CTE1 a
LEFT OUTER JOIN CTE2 b ON a.station_id = b.station_id
WHERE a.tcl_total - isnull(b.total,0) <> 0

【讨论】：

【解决方案3】：

SELECT a.station_id AS station, a.tcl_total-b.total AS difference
(
 SELECT station_id, SUM(tcl_missing + tcl_not_missing) as tcl_total
 FROM tcl_missing_summary 
 GROUP BY station_id
) as a 
INNER JOIN
(
 SELECT station_id, SUM(total) as total
 FROM tcl_breakdown_op
 WHERE tr_standard not like '%cru'
 GROUP BY station_id
) as b ON a.station_id = b.station_id
 WHERE a.tcl_total - b.total <> 0

如果您知道一个结果集可能没有站点的某些子集的数据，也可以将其转换为外连接。

【讨论】：

【解决方案4】：

我同意 rs 对 CTE 方法的看法，因为嵌套选择恕我直言，在其中两个或三个之后会变得讨厌。 CTE 还一次按一组逻辑地显示您的数据。

但是，如果您必须使用嵌套选择，只需添加几行即可使其完整：

SELECT a.tcl_total
from 
(
 SELECT station_id, SUM(tcl_missing + tcl_not_missing) as tcl_total
 FROM tcl_missing_summary 
 GROUP BY station_id
) as a 
,
(
 SELECT station_id, SUM(total) as total
 FROM tcl_breakdown_op
 WHERE tr_standard not like '%cru'
 GROUP BY station_id
) as b
 WHERE a.tcl_total - b.total <> 0

【讨论】：