正如标题所示,我想选择用GROUP BY分组的每组行中的第一行。

具体来说,如果我有一个如下所示的采购表:

SELECT * FROM purchases;

我的输出:

id customer total
1 Joe 5
2 Sally 3
3 Joe 2
4 Sally 1

我想查询每个客户的最大购买量(总购买量)。类似于:

SELECT FIRST(id), customer, FIRST(total)
FROM  purchases
GROUP BY customer
ORDER BY total DESC;

预期输出:

FIRST(id) customer FIRST(total)
1 Joe 5
2 Sally 3

当前回答

Snowflake/Teradata支持QUALIFY子句,其工作方式与窗口函数的HAVING类似:

SELECT id, customer, total
FROM PURCHASES
QUALIFY ROW_NUMBER() OVER(PARTITION BY p.customer ORDER BY p.total DESC) = 1

其他回答

如果要从聚合行集合中选择任何行(根据特定条件)。如果您想使用除max/min之外的另一个(sum/avg)聚合函数。因此,您不能在DISTINCT ON时使用线索

您可以使用下一个子查询:

SELECT  
    (  
       SELECT **id** FROM t2   
       WHERE id = ANY ( ARRAY_AGG( tf.id ) ) AND amount = MAX( tf.amount )   
    ) id,  
    name,   
    MAX(amount) ma,  
    SUM( ratio )  
FROM t2  tf  
GROUP BY name

您可以将amount=MAX(tf.amount)替换为任何需要的条件,但有一个限制:此子查询不能返回多行

但是如果你想做这样的事情,你可能需要寻找窗口函数

对PostgreSQL、U-SQL、IBM DB2和Google BigQuery SQL使用ARRAY_AGG函数:

SELECT customer, (ARRAY_AGG(id ORDER BY total DESC))[1], MAX(total)
FROM purchases
GROUP BY customer

这是我们如何通过使用windows函数实现的:

    create table purchases (id int4, customer varchar(10), total integer);
    insert into purchases values (1, 'Joe', 5);
    insert into purchases values (2, 'Sally', 3);
    insert into purchases values (3, 'Joe', 2);
    insert into purchases values (4, 'Sally', 1);
    
    select ID, CUSTOMER, TOTAL from (
    select ID, CUSTOMER, TOTAL,
    row_number () over (partition by CUSTOMER order by TOTAL desc) RN
    from purchases) A where RN = 1;

这可以通过MAX FUNCTION on total和GROUP by id和customer轻松实现。

SELECT id, customer, MAX(total) FROM  purchases GROUP BY id, customer
ORDER BY total DESC;

在支持CTE和窗口功能的数据库上:

WITH summary AS (
    SELECT p.id, 
           p.customer, 
           p.total, 
           ROW_NUMBER() OVER(PARTITION BY p.customer 
                                 ORDER BY p.total DESC) AS rank
      FROM PURCHASES p)
 SELECT *
   FROM summary
 WHERE rank = 1

任何数据库都支持:

但你需要添加逻辑来打破联系:

  SELECT MIN(x.id),  -- change to MAX if you want the highest
         x.customer, 
         x.total
    FROM PURCHASES x
    JOIN (SELECT p.customer,
                 MAX(total) AS max_total
            FROM PURCHASES p
        GROUP BY p.customer) y ON y.customer = x.customer
                              AND y.max_total = x.total
GROUP BY x.customer, x.total