SQL如何按状态选择交易金额最高的客户

Question

我正在尝试编写一个 SQL 查询，返回每个州花费最多的五个客户的姓名和购买金额。

表架构

customers
|_state
|_customer_id
|_customer_name

transactions
|_customer_id
|_transact_amt

尝试看起来像这样

SELECT state, Sum(transact_amt) AS HighestSum
FROM (
    SELECT name, transactions.transact_amt, SUM(transactions.transact_amt) AS HighestSum
    FROM customers
    INNER JOIN customers ON transactions.customer_id = customers.customer_id
    GROUP BY state
) Q
GROUP BY transact_amt
ORDER BY HighestSum 

我迷路了。谢谢。

预期结果是每个州交易量最高的前 5 名客户的姓名。

ERROR:  table name "customers" specified more than once
SQL state: 42712

Answer 1

首先，您的JOIN 需要正确。二、你要使用窗口函数：

SELECT ct.*
FROM (SELECT c.customer_id, c.name, c.state, SUM(t.transact_amt) AS total,
             ROW_NUMBER() OVER (PARTITION BY c.state ORDER BY SUM(t.transact_amt) DESC) as seqnum
      FROM customers c JOIN
           transaactions t
           ON t.customer_id = c.customer_id
      GROUP BY c.customer_id, c.name, c.state
     ) ct
WHERE seqnum <= 5;

您的 SQL 似乎有几个问题。我将从了解聚合函数开始。您有一个别名为 HighestSum 的 SUM()。这只是每位客户的总数。

Answer 2

您可以使用聚合获取它们，然后使用RANK() 窗口函数。例如：

select
  state,
  rk,
  customer_name
from (
  select
    *,
    rank() over(partition by state order by total desc) as rk
  from (
    select
      c.customer_id,
      c.customer_name,
      c.state,
      sum(t.transact_amt) as total
    from customers c
    join transactions t on t.customer_id = c.customer_id
    group by c.customer_id
  ) x
) y
where rk <= 5
order by state, rk

Answer 3

已经有两个有效的答案。这是第三个：

SELECT *
FROM  (
   SELECT c.state, c.customer_name, t.*
        , row_number() OVER (PARTITION BY c.state ORDER BY t.transact_sum DESC NULLS LAST, customer_id) AS rn
   FROM  (
      SELECT customer_id, sum(transact_amt) AS transact_sum
      FROM   transactions
      GROUP  BY customer_id
      ) t
   JOIN   customers c USING (customer_id)
   ) sub
WHERE  rn < 6
ORDER  BY state, rn;

要点

• 当聚合一个大表的所有或大部分行时，聚合之前通常快得多。假设参照完整性（FK 约束），我们不会聚合否则会被过滤的行。当加入更多聚合表时，这可能会从“必备”变为纯粹的必需。相关：

  • Why does the following join increase the query time significantly?
  • Two SQL LEFT JOINS produce incorrect result

• 在窗口函数中添加额外的ORDER BY 项目以定义要从关联中选择哪些行。在我的示例中，它只是customer_id。如果您没有决胜局，则在平局的情况下结果是任意的，这可能没问题。但是每一次其他执行可能会返回不同的结果，这通常是一个问题。。 或者您在结果中包含所有平局。然后我们回到rank() instead of row_number()。见：

  • PostgreSQL equivalent for TOP n WITH TIES: LIMIT "with ties"?

• 虽然transact_amt 可以为 NULL（尚未排除），但任何总和最终也可能为 NULL。在毫无戒心的ORDER BY t.transact_sum DESC 中，这些客户名列前茅，因为 NULL 按降序排列在第一位。使用 DESC NULLS LAST 来避免这个陷阱。 （或将transact_amt 列定义为NOT NULL。）

  • PostgreSQL sort by datetime asc, null first?