如何为MySQL中的每个组选择第一行？

Question

在C#中它会是这样的:

table
   .GroupBy(row => row.SomeColumn)
   .Select(group => group
       .OrderBy(row => row.AnotherColumn)
       .First()
   )

Linq-To-Sql将其转换为以下T-SQL代码:

SELECT [t3].[AnotherColumn], [t3].[SomeColumn]
FROM (
    SELECT [t0].[SomeColumn]
    FROM [Table] AS [t0]
    GROUP BY [t0].[SomeColumn]
    ) AS [t1]
OUTER APPLY (
    SELECT TOP (1) [t2].[AnotherColumn], [t2].[SomeColumn]
    FROM [Table] AS [t2]
    WHERE (([t1].[SomeColumn] IS NULL) AND ([t2].[SomeColumn] IS NULL))
      OR (([t1].[SomeColumn] IS NOT NULL) AND ([t2].[SomeColumn] IS NOT NULL)
        AND ([t1].[SomeColumn] = [t2].[SomeColumn]))
    ORDER BY [t2].[AnotherColumn]
    ) AS [t3]
ORDER BY [t3].[AnotherColumn]

但它与MySQL不兼容.

Answer 1

我的答案只基于你帖子的标题,因为我不知道C#并且不理解给定的查询.但是在MySQL中,我建议你尝试子选择.首先获取一组有趣列的主键,然后从这些行中选择数据:

SELECT somecolumn, anothercolumn 
  FROM sometable 
 WHERE id IN (
               SELECT min(id) 
                 FROM sometable 
                GROUP BY somecolumn
             );

Answer 2

当我写作

SELECT AnotherColumn
FROM Table
GROUP BY SomeColumn
;

有用.IIRC在其他RDBMS中这样的语句是不可能的,因为不属于分组键的列被引用而没有任何类型的聚合.

这种"怪癖"与我想要的非常接近.所以我用它来得到我想要的结果:

SELECT * FROM 
(
 SELECT * FROM `table`
 ORDER BY AnotherColumn
) t1
GROUP BY SomeColumn
;

Answer 3

这是您可以尝试的另一种方式,不需要该ID字段.

select some_column, min(another_column)
  from i_have_a_table
 group by some_column

我仍然同意lfagundes你应该添加一些主键..

还要注意,通过执行此操作,您不能(轻松)获取其他值与生成的some_colum,another_column对相同的行!你需要lfagundes apprach和PK来做到这一点!

Answer 4

您应该使用一些聚合函数来获取所需的AnotherColumn的值.也就是说,如果您想为SomeColumn的每个值(数字或词典)获得AnotherColumn的最低值,您可以使用:

SELECT SomeColumn, MIN(AnotherColumn)
FROM YourTable
GROUP BY SomeColumn

一些希望有用的链接:

http://dev.mysql.com/doc/refman/5.1/en/group-by-functions.html

http://www.oreillynet.com/databases/blog/2007/05/debunking_group_by_myths.html

Answer 5

SELECT
    t1.*

FROM
    table_name AS t1

    LEFT JOIN table_name AS t2 ON (
        t2.group_by_column = t1.group_by_column
        -- group_by_column is the column you would use in the GROUP BY statement
        AND
        t2.order_by_column < t1.order_by_column
        -- order_by_column is column you would use in the ORDER BY statement
        -- usually is the autoincremented key column
    )

WHERE
    t2.group_by_column IS NULL;

使用 MySQL v8+，您可以使用窗口函数

Answer 6

我建议使用 MySql 的官方方式：

SELECT article, dealer, price
FROM   shop s1
WHERE  price=(SELECT MAX(s2.price)
              FROM shop s2
              WHERE s1.article = s2.article
              GROUP BY s2.article)
ORDER BY article;

这样我们就可以得到每件物品的最高价格

Answer 7

来自MySQL 5.7 文档

MySQL 5.7.5 及更高版本实现了功能依赖检测。如果启用了 ONLY_FULL_GROUP_BY SQL 模式（默认情况下是这样），MySQL 将拒绝选择列表、HAVING 条件或 ORDER BY 列表引用非聚合列的查询，这些列既不在 GROUP BY 子句中命名，也不在功能上依赖它们.

这意味着@Jader Dias 的解决方案不会在任何地方都有效。

这是一个在ONLY_FULL_GROUP_BY启用时有效的解决方案：

SET @row := NULL;
SELECT
    SomeColumn,
    AnotherColumn
FROM (
    SELECT
        CASE @id <=> SomeColumn AND @row IS NOT NULL 
            WHEN TRUE THEN @row := @row+1 
            ELSE @row := 0 
        END AS rownum,
        @id := SomeColumn AS SomeColumn,
        AnotherColumn
    FROM
        SomeTable
    ORDER BY
        SomeColumn, -AnotherColumn DESC
) _values
WHERE rownum = 0
ORDER BY SomeColumn;

Answer 8

我在答案中没有看到以下解决方案，所以我想我会把它放在那里。

问题是AnotherColumn在按 分组的所有组中按顺序选择第一行SomeColumn。

以下解决方案将在 MySQL 中执行此操作。id必须是一个唯一的列，它不能包含包含-（我用作分隔符）的值。

select t1.*
from mytable t1
inner join (
  select SUBSTRING_INDEX(
    GROUP_CONCAT(t3.id ORDER BY t3.AnotherColumn DESC SEPARATOR '-'),
    '-', 
    1
  ) as id
  from mytable t3
  group by t3.SomeColumn
) t2 on t2.id = t1.id


-- Where 
SUBSTRING_INDEX(GROUP_CONCAT(id order by AnotherColumn desc separator '-'), '-', 1)
-- can be seen as:
FIRST(id order by AnotherColumn desc)

-- For completeness sake:
SUBSTRING_INDEX(GROUP_CONCAT(id order by AnotherColumn desc separator '-'), '-', -1)
-- would then be seen as:
LAST(id order by AnotherColumn desc)

有一个功能要求的FIRST()，并LAST()在MySQL的bug跟踪系统，但它被关闭多年了。

Answer 9

最佳性能且易于使用：

SELECT id, code,
SUBSTRING_INDEX( GROUP_CONCAT(price ORDER BY id DESC), ',', 1) first_found_price
FROM stocks
GROUP BY code
ORDER BY id DESC

Answer 10

rtribaldos 提到，在较新的数据库版本中，可以使用窗口函数。\n这是一个对我有用的代码，并且与 Martin Zwar\xc3\xadk 的substring_index -解决方案（在 Mariadb 10.5.16 中）
一样快：

\n

SELECT group_col, order_col FROM (\n  SELECT group_col, order_col\n  , ROW_NUMBER() OVER(PARTITION BY group_col ORDER BY order_col) rnr \n  FROM some_table\n  WHERE <some_condition>\n) i\nWHERE rnr=1;\n

\n