Question

我有这张桌子

 |      date    |  sum  |
 |--------------|-------|
 | 2015-02-19   | 10000 |
 | 2015-02-19   | 10000 |
 | 2015-02-20   | 15000 |
 | 2015-02-20   | 15000 |
 | 2015-02-21   | 18000 |
 | 2015-02-21   | 18000 |

我想从表中选择前2行，但只选择不同的行，这意味着我的结果应该返回2015-02-20和2015-02-21。

SELECT TOP 2 distinct date 
FROM stock

使用它会给我一个错误：

关键字＆＃39; distinct＆＃39;附近的语法不正确。

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Answer 1

您可以尝试这样

select top 2 * from
(
    select distinct date FROM stock
)

Answer 2

尝试类似：

SELECT TOP 2 date 
FROM stock
GROUP BY date

Answer 3

我认为Distinct和Top应该在您的查询中切换位置：

SELECT DISTINCT TOP 2 date FROM stock ORDER BY date DESC

Answer 4

试

   select distinct top 2 date  from stock

Answer 5

您可以使用GROUP BY：

SELECT TOP 2 date 
FROM stock
GROUP BY date
ORDER BY date DESC

示例结果：

DATE
2015-02-21
2015-02-20

请参阅SQL Fiddle中的结果。

Answer 6

试试这个：

  WITH cte AS
 ( SELECT distinct date ,
       ROW_NUMBER() OVER (PARTITION BY date
                          ORDER BY date DESC
                         )
         AS rn
   FROM stock
 )
 SELECT date
 FROM cte
 WHERE rn <= 3
 ORDER BY rn ;

Answer 7

试试这个：

SELECT TOP 2 date FROM stock group by date

选择彼此不同的前2行

7 个答案: