Question

好吧我有一张表，我需要从GROUP BY函数中获取相关日期......

Here is a sample of the data:
min_price, day
140000     2010-09-26
130000     2010-09-27
154991     2010-10-08
143000     2010-10-09
156470     2010-10-10

I would like this result:
130000     2010-09-27
143000     2010-10-09

Here is the SQL I use:

SELECT MIN(min_price) AS low_price, min_price, day
FROM compress
WHERE item_name = '$item'
GROUP BY EXTRACT(MONTH FROM day), EXTRACT(YEAR FROM day)

Here is what I get:
130000     2010-09-26
143000     2010-10-08

如您所见，日期与最小行值不匹配。我需要他们匹配。我可以用什么SQL来获得结果？

Answer 1

Select C.min_price, C.day
From compress As C
Where C.item_name = '$item'
    And Exists  (
                Select 1
                From compress As C2
                Where C2.item_name = C.item_name
                    And Extract(Month From C2.day) = Extract(Month From C.day)
                    And Extract(Year From C2.day) = Extract(Year From C.day)
                Group By Extract(Month From C2.day), Extract(Year From C2.day)
                Having C.min_price = Min(C2.min_price)
                )

<强> ADDITION

MySQL更适合加入，所以这里是同一查询的变体

Select C.min_price, C.day
From compress As C
    Join    (
            Select C2.item_name
                , Extract(Month From C2.day) As ItemMonth
                , Extract(Year From C2.day) As ItemYear
                , Min(C2.min_price) As MinPrice
            From compress As C2
            Group By C2.item_num, Extract(Month From C2.day), Extract(Year From C2.day)
            ) As Z
        On Z.item_name = C.item_name
            And Z.ItemMonth = Extract(Month From C.day)
            And Z.ItemYear = Extract(Year From C.day)
            And Z.MinPrice = C.min_price
Where C.item_name = '$item'

真正放慢任何变化的是每行调用Extract。如果您的日历表包含每个日期的行以及日期的月份和年份的列，并且您的日历表涵盖了数据中的日期范围，那么您可以执行以下操作：

Select C.min_price, C.day
From compress As C
    Join Calendar As Cal
        On Cal.Date = C.Day
    Join    (
            Select C3.item_name
                , Cal3.Month, Cal3.Year
                , Min(C3.min_price) As MinPrice
            From compress As C3
                Join Calendar As Cal3
                    On Cal3.Date = C3.day
            Group By C3.item_num, Cal3.Month, Cal3.Year
            ) As Z
        On Z.item_name = C.Item_name
            An Z.Month = Cal.Month
            And Z.Year = Cal.Year
            And Z.MinPrice = C.min_price
Where C.item_name = '$item'

Answer 2

SELECT min_price AS low_price, min_price, day
FROM compress
WHERE item_name = '$item' AND min_price = MIN(min_price)
GROUP BY EXTRACT(MONTH FROM day), EXTRACT(YEAR FROM day)

然后，这将选择min_price与MIN（min_price）

相同的行

Answer 3

您的SQL无效。在分组时，显示列表中的每一列（不是表达式的组的一部分）必须通过聚合函数传递，例如， AVG（），MIN（），MAX（），但你只是要求“白天”。

请用自然语言（英语）说明您想要达到的目标。你的例子不是很清楚。

Answer 4

如果我已正确理解问题，这应该有效......

mysql> select * from stack;
+-----------+------------+
| min_price | myday      |
+-----------+------------+
|    140000 | 2010-09-26 | 
|    130000 | 2010-09-27 | 
|    154991 | 2010-10-08 | 
|    143000 | 2010-10-09 | 
|    156470 | 2010-10-10 | 
+-----------+------------+
5 rows in set (0.00 sec)

mysql> select a.* from stack as a inner join 
    -> (select min(min_price) as my_price from stack group by extract(YEAR_MONTH from myday)) as b
    -> on a.min_price = b.my_price;
+-----------+------------+
| min_price | myday      |
+-----------+------------+
|    130000 | 2010-09-27 | 
|    143000 | 2010-10-09 | 
+-----------+------------+
2 rows in set (0.00 sec)

您需要加入主键才能避免意外结果。如果查询速度很慢，请为subselect创建一个临时表，在其上构建索引，然后使用该表与主表联接。

SQL MIN（值）匹配日期

4 个答案: