Question

我有以下查询。我需要在我的组中仅选择最旧的行（按DATE列）。保持我的条件（在哪里和分组）。

SELECT 
table.*
from table
where views > $views AND sales>23
group by 
name

更清楚一点就像传递frm这第一张桌子

+--------------------+-----------+--------------------+-----------+
| name               | views     | DATE               |   sales   |
+--------------------+-----------+--------------------+-----------+
| sue1               |       494 | 2014-06-23 12:08:29|   26      |
| sue2               |       494 | 2014-06-25 12:08:29|   34      |
| sue3               |       494 | 2014-06-27 12:08:29|   45      |
| sue4               |       520 | 2014-06-26 12:08:29|   56      |
| sue5               |       570 | 2014-06-24 12:08:29|   20      |
| sue5               |       570 | 2014-06-24 12:08:29|   28      |
+--------------------+-----------+--------------------+-----------+

到第二个

+--------------------+-----------+--------------------+-----------+
|  name              | views     | DATE               |  sales    |
+--------------------+-----------+--------------------+-----------+
| sue3               |       494 | 2014-06-27 12:08:29|    45     |
| sue4               |       520 | 2014-06-26 12:08:29|    56     |
| sue5               |       570 | 2014-06-24 12:08:29|    28     |
+--------------------+-----------+--------------------+-----------+

marko.c

我尝试过做

的子选择

SELECT 
table.*
from table
where views > $views AND sales>23
and date in (select max(date) from table where views > $views AND sales>23)
group by 
name

但它没有用。任何人

Answer 1

作为第一关，老实说，我不确定你想要什么。我认为您的要求是按照最早的日期获取所有独特的观看次数...但无论如何，如果您想要每件事物的最早日期，那么只需在订购后再进行分组。

SELECT * FROM(
    SELECT 
    table.*
    FROM table
    WHERE views > $views AND sales>23
    ORDER BY date
) AS t
GROUP BY name;

当您在某个日期进行聚合时，请将其反向思考..

MAX(date) will give you the NEWEST date because it is the greatest date. 
MIN(date) will give you the OLDEST date because it is the oldest / smallest date.

所以如果你想做其他发布的查询，你可以做

SELECT 
table.*
from table
where views > $views AND sales>23
and date in (select MIN(date) from table where views > $views AND sales>23)
group by 
name

并查看MIN（日期）是否可以解决问题

Answer 2

您正在选择子查询中的最大总日期，但实际需要的是每个名称的最小（最早）日期。

SELECT 
    t1.*
from table t1
join (
    select 
        t2.name,
        min(t2.date) min_date
    from table t2
    where t2.views > $views AND t2.sales > 23
    group by t2.name
) t2 on t2.name = t1.name and t2.min_date = t1.date
where t1.views > $views AND t1.sales > 23

mysql where a group in group by statement？

2 个答案: