0

这可能之前已经讨论过(或者太简单了),但我找不到一个简单的答案:应该如何完成:选择(单个)最早的未来日期和每个 id 的最近日期?

t1:             ==?==>         (earliest)  (most recent)
|id|date(DESC)|          |id|future_date  |   past_date   |
+==+==========+          +==+=============+===============+
|1 |  d1      |          | 1|       d1    |        d3     |
|2 |  d2      |          | 2|       d2    |        d6     |
           (<==now)      | 3|             |        d4     |
|1 |  d3      |
|3 |  d4      |
|1 |  d5      |
|2 |  d6      |

我正在按照以下思路进行思考,但我觉得这是相当复杂/糟糕的语法,而且我还没有弄清楚如何限制单个结果(即最近/将来最近)。有什么建议么?

 SELECT t_1.id,t_1.date AS future_date,t_2.date AS past_date 
    FROM (SELECT * FROM t1 WHERE t1.date>CURRENT_TIMESTAMP) t_1
    LEFT OUTER JOIN 
    (SELECT TOP 1 * FROM t1 WHERE t1.date<CURRENT_TIMESTAMP) t_2 
    ON t_1.id=t_2.id
4

2 回答 2

0

我想你可以使用:

   SELECT x.id,
          MIN(y.date) AS future_date,
          MAX(z.date) AS past_date
     FROM (SELECT DISTINCT t.id
             FROM YOUR_TABLE t) x
LEFT JOIN YOUR_TABLE y ON y.id = x.id
                      AND y.date > CURRENT_TIMESTAMP
LEFT JOIN YOUR_TABLE z ON z.id = x.id
                      AND z.date < CURRENT_TIMESTAMP
 GROUP BY x.id

目前尚不清楚您正在使用什么数据库 - TOP 仅是 TSQL/SQL Server 语法(2000+),而 NOWNOW()由 MySQL 和 PostgreSQL 支持...... CURRENT_TIMESTAMP 是 ANSI,并且被所有人支持。

于 2011-01-20T07:18:19.087 回答
0

为什么这行不通?

select min(date) past_date,max(DATE)future_date
from t1
HAVING MIN(date) < CURRENT_TIMESTAMP AND MAX(DATE) < CURRENT_TIMESTAMP
group by id
于 2011-01-20T08:12:15.067 回答