SQL server中其他记录中剩余记录之和的前10条记录

我有一个表，它由4列组成-（id int，location varchar（max），[date]datetime，Total_Sales Money），其中包括输出中的20条记录。我想要11行，其中前10行是原样，第11行应该是剩余行的总和，即其他_位置的11到20行。请帮我查询一下

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我想要11行，其中前10行按原样排列，第11行应该是剩余行的总和

您可以使用

union

选择前10行和1行总和

可能是代码：

SELECT id, location, date, total_sales FROM table limit 10 // where etc..
UNION
SELECT 0 as id, '' as location, 0 as date, sum(total_sales) as total_sales FROM table;

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在SQL Server 2012中，使用

ROW\u NUMBER（）

window函数可以实现这一点。创建

ROW\u NUMBER（）

为每行设置

RowNum

，并基于结果，使用

UNION ALL

返回结果：

;WITH CTE AS (
    SELECT id, location, [date], Total_Sales,
           ROW_NUMBER() OVER (ORDER BY Id) AS RowNum
    FROM TableName
)

SELECT CAST(id AS VARCHAR(30)) AS Id,
       location, 
       CAST([date] AS VARCHAR(30)) AS [date], 
       CAST(Total_Sales AS VARCHAR(30)) AS Total_Sales 
FROM CTE 
WHERE RowNum BETWEEN 1 AND 10
UNION 
SELECT '11-20 ids' AS id, 
       '11-20 AS locations' as location, 
       '11-20 dates' AS [date], 
       SUM(Total_Sales) AS Total_Sales 
FROM CTE 
WHERE RowNum BETWEEN 11 AND 20

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我会使用

行数（）

和条件聚合。这样，您就可以使用一个

分组方式来执行此操作：

select (case when seqnum <= 10 then id end) as id,
       max(case when seqnum <= 10 then location else 'All else' end) as location,
       max(case when seqnum <= 10 then datetime end) as datetime,
       sum(amount)
from (select t.*,
             row_number() over (order by amount desc) as seqnum
      from t
     ) t
group by (case when seqnum <= 10 then id end);

select（seqnum时的情况）如何定义“前10行”？按日期？按金额？