我将假设同一个ID在同一天总是有相同的开销。我还将假设您的RDBMS支持派生表。在这种情况下，您需要的是：

select date, sum(cost)
from 
  (select distinct date, id, cost from YourTable)
group by date

已更新

Oracle派生表不需要别名。

I think all you want to do here is group by both date and id, as MPelletier wrote.
FWIW - you can do a distinct inside an aggregate in some dbs: http://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html#function_sum
And even better, use group by with rollup to get totals along with the grouping: http://dev.mysql.com/doc/refman/5.0/en/group-by-modifiers.html

SELECT DATE, ID, SUM(cost)
FROM table_name
GROUP BY DATE, ID WITH ROLLUP;

我认为张贴问题的逻辑有一些缺陷。
假设您有以下结果集：

DATE          ID    cost    
---------------------------------------
 01/01/2011    1     10      
 01/01/2011    1     15     
 01/01/2011    2     10      
 01/01/2011    2     10

查询应返回什么？

DATE          cost    
---------------------------------------
 01/01/2011    20

或

DATE          cost    
---------------------------------------
 01/01/2011    25

接受的答案将返回35。
还是说这样的一套根本不可能有？

不了解数据库/SQL引擎：
SELECT日期，SUM（成本）FROM您的表GROUP BY日期，ID;

SELECT Date, First(TCost)
FROM
(SELECT Date, Sum(Cost) as TCost FROM YourTable GROUP BY Date, ID)
GROUP BY Date

您可以使用此查询。

SELECT dates, Sum(cost) as TCost FROM TABLE_NAME GROUP BY  ID

也许你可以用这个黑客。
将您要加总的值-cost-与id字段连接起来，并在它们之间插入一个字母，例如'a'。您可以将DISTINCT用于这些连接的字段，以便在聚合函数中只使用一次。
当你使用SUM时，这些字符串字段自动转换为整数，MySQL会在第一个字母处将其截断。
这是因为有许多不同的字符串可以转换为值1，例如“1”、“1”或“1a”。
来源：https://dev.mysql.com/doc/refman/5.7/en/type-conversion.html

SELECT
    id,
    SUM(cost),
    SUM(DISTINCT cost),
    GROUP_CONCAT(CONCAT(cost, 'a', id)),
    GROUP_CONCAT(DISTINCT CONCAT(cost, 'a', id)),
    SUM(DISTINCT CONCAT(cost, 'a', id)) expectedSum
FROM (
    SELECT '2011-01-01' AS date, 1 AS id, 10 AS cost
    UNION ALL SELECT '2011-01-01', 1, 10
    UNION ALL SELECT '2011-01-01', 2, 10
    UNION ALL SELECT '2011-01-01', 2, 10
) __tmp
GROUP BY
    id

小心！这不是一个很好的通用SQL解决方案，这是一个技巧。我不确定它是否适用于每个特殊情况。

试试这个

SELECT SUM(cost) AS Expr1
     , ID
     , date
 FROM (SELECT DISTINCT date
            , ID
            , cost
         FROM table_name AS table_name_1
        GROUP BY date, ID, cost) AS derivedtbl_1
 GROUP BY ID, date

或此

SELECT     SUM(cost) AS Expr1,  date
  FROM ( SELECT DISTINCT date, ID, cost
           FROM table_name AS table_name_1
           GROUP BY date, ID, cost) AS derivedtbl_1
 GROUP BY date

试试这个

我正在使用Mysql进行此. DEMO TABLE（表名为“tablename”. id，成本类型为“int”）

id | cost  
 1 | 15  
 1 | 25  
 1 | 10  
 2 | 25  
 2 | 5  
 2 | 10  
 3 | 20    
 3 | 15    
 3 | 10    
 4 | 20    
 4 | 20

查询使用

select id
     , SUM(cost) as 'Total Cost' 
  from sumchk 
 group by id

或（使用下面的一个）

select id, sum(cost) as 'Total Cost'  
  from (select * 
          from tablename
  ) a  
group by id;

结果：

id  |  Total Cost  
   1  |    50  
   2  |    40  
   3  |    45  
   4  |    40