Question

我正在多个记录上执行聚合函数，这些记录按公共ID分组。问题是，我还想导出一些其他字段，这些字段在分组记录中可能不同，但我想从其中一个记录中获取这些字段（第一个，根据查询的ORDER BY）。

起点示例：

SELECT
  customer_id,
  sum(order_total),
  referral_code
FROM order
GROUP BY customer_id
ORDER BY date_created

我需要查询引用代码，但是在聚合函数之外执行它意味着我必须按该字段进行分组，这不是我想要的 - 在这个示例中我需要每个客户一行。我真的只关心第一个订单的推荐代码，我很乐意抛弃任何后来的推荐代码。

这是在PostgreSQL中，但也许来自其他数据库的语法可能足够相似。

拒绝解决方案：

无法使用max（）或min（），因为订单很重要。
子查询可能最初有效，但不能扩展;这是一个极其简化的例子。我的实际查询有几十个字段，比如referral_code，我只想要第一个实例，以及几十个WHERE子句，如果在子查询中重复，会导致维护噩梦。

Answer 1

嗯，实际上非常简单。

首先，让我们编写一个将进行聚合的查询：

select customer_id, sum(order_total)
from order
group by customer_id

现在，让我们编写一个查询，返回给定customer_id的1st referral_code和date_created：

select distinct on (customer_id) customer_id, date_created, referral_code
from order
order by customer_id, date_created

现在，您只需加入2个选项：

select
    x1.customer_id,
    x1.sum,
    x2.date_created,
    x2.referral_code
from
    (
        select customer_id, sum(order_total)
        from order
        group by customer_id
    ) as x1
    join
    (
        select distinct on (customer_id) customer_id, date_Created, referral_code
        from order
        order by customer_id, date_created
    ) as x2 using ( customer_id )
order by x2.date_created

我没有对它进行测试，因此可能存在拼写错误，但通常它应该可以正常工作。

Answer 2

您需要window functions。这是GROUP BY，但您仍然可以访问各行。虽然只使用了Oracle等价物。

Answer 3

也许是这样的：

SELECT
     O1.customer_id,
     O1.referral_code,
     SQ.total
FROM
     Orders O1
LEFT OUTER JOIN Orders O2 ON
     O2.customer_id = O1.customer_id AND
     O2.date_created < O1.date_created
INNER JOIN (
     SELECT
          customer_id,
          SUM(order_total) AS total
     FROM
          Orders
     GROUP BY
          customer_id
     ) SQ ON SQ.customer_id = O1.customer_id
WHERE
     O2.customer_id IS NULL

Answer 4

如果date_created保证每个customer_id是唯一的，那么你可以这样做：

[简单表]

create table ordertable (customer_id int, order_total int, referral_code char, date_created datetime)
insert ordertable values (1,10, 'a', '2009-01-01')
insert ordertable values (2,15, 'b', '2009-01-02')
insert ordertable values (1,35, 'c', '2009-01-03')

[用更好的东西替换我的蹩脚表格名称：]]

SELECT
  orderAgg.customer_id,
  orderAgg.order_sum,
  referral.referral_code as first_referral_code
FROM (
        SELECT
          customer_id,
          sum(order_total) as order_sum
        FROM ordertable
        GROUP BY customer_id
    ) as orderAgg join (
        SELECT
          customer_id,
          min(date_created) as first_date
        FROM ordertable
        GROUP BY customer_id
    ) as dateAgg on orderAgg.customer_id = dateAgg.customer_id
    join ordertable as referral 
        on dateAgg.customer_id = referral.customer_id
            and dateAgg.first_date = referral.date_created

Answer 5

这样的事情能做到吗？

SELECT
  customer_id,
  sum(order_total),
  (SELECT referral_code 
   FROM order o 
   WHERE o.customer_id = order.customer_id 
   ORDER BY date_created 
   LIMIT 1) AS customers_referral_code
FROM order
GROUP BY customer_id, customers_referral_code
ORDER BY date_created

这不需要你在两个地方维护WHERE子句并保持顺序重要性，但如果你需要像referral_code这样的“几十个字段”，它会变得非常毛茸茸。它也相当慢（至少在MySQL上）。

听起来像referral_code，而且像它一样的数十个字段应该在客户表中，而不是订单表，因为它们在逻辑上与客户1：1关联，而不是订单。将它们移动到那里会使查询更简单。

这也可以解决问题：

SELECT
  o.customer_id,
  sum(o.order_total),
  c.referral_code, c.x, c.y, c.z
FROM order o LEFT JOIN (
    SELECT referral_code, x, y, z
    FROM orders c 
    WHERE c.customer_id = o.customer_id 
    ORDER BY c.date_created
    LIMIT 1
) AS c
GROUP BY o.customer_id, c.referral_code
ORDER BY o.date_created

Answer 6

SELECT  customer_id, order_sum,
        (first_record).referral, (first_record).other_column
FROM    (
        SELECT  customer_id,
                SUM(order_total) AS order_sum,
                (
                SELECT  oi
                FROM    order oi
                WHERE   oi.customer_id = o.customer_id
                LIMIT 1
                ) AS first_record
        FROM    order o
        GROUP BY
                customer_id
        ) q

对相关记录进行分组，但仅从第一条记录中选择某些字段

6 个答案: