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
我通过窗口函数dbfiddle的方法：
将每组的row_number（）分配给（按agreement_id、order_id划分）为nrow只取组：filter中的第一行（其中nrow=1）
with intermediate as (select
*,
row_number() over ( partition by agreement_id, order_id ) as nrow,
(sum( suma ) over ( partition by agreement_id, order_id ))::numeric( 10, 2) as order_suma,
from <your table>)
select
*,
sum( order_suma ) filter (where nrow = 1) over (partition by agreement_id)
from intermediate
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
在PostgreSQL中，另一种可能是将first_value窗口函数与SELECT DISTINCT结合使用：
select distinct customer_id,
first_value(row(id, total)) over(partition by customer_id order by total desc, id)
from purchases;
我创建了一个组合（id，total），因此两个值都由同一个聚合返回。当然，您可以始终应用first_value（）两次。
非常快速的解决方案
SELECT a.*
FROM
purchases a
JOIN (
SELECT customer, min( id ) as id
FROM purchases
GROUP BY customer
) b USING ( id );
如果表是按id索引的，则速度非常快：
create index purchases_id on purchases (id);
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
对于SQl Server，最有效的方法是：
with
ids as ( --condition for split table into groups
select i from (values (9),(12),(17),(18),(19),(20),(22),(21),(23),(10)) as v(i)
)
,src as (
select * from yourTable where <condition> --use this as filter for other conditions
)
,joined as (
select tops.* from ids
cross apply --it`s like for each rows
(
select top(1) *
from src
where CommodityId = ids.i
) as tops
)
select * from joined
不要忘记为使用过的列创建聚集索引
