MySQL 查找未首次购买的客户的交易（付款）答案

【问题标题】：MySQL find transactions (payments) of customers that have not purchased for the first timeMySQL 查找未首次购买的客户的交易（付款）
【发布时间】：2015-07-29 19:03:13
【问题描述】：

我有一个包含交易的简单表格，我想了解每个月有多少消费者进行了总计超过 0 的交易，并且他们的第一笔交易不在该月。首次交易是指客户在当月第一次购买。

我想要得到的结果如下：

+--------+---------+-----------------------------------+
|  Year  |  Month  |  NumOfCustomersWithPositiveTotals |
+--------+----------------------------+----------------+
|  2014  |    1    |                 22                |
+--------+----------------------------+----------------+
|  2014  |    2    |                 10                |
+--------+----------------------------+----------------+

我有一个 SQL 小提琴，我可以在其中找到相同的东西，但适用于当月内完成第一笔交易的消费者。实际上，我正在寻找的查询是相同的，但对于其他消费者而言。

这是小提琴：http://sqlfiddle.com/#!9/31538/24

【问题讨论】：

标签： mysql payment

【解决方案1】：

我想这就是你想要的：

SELECT count(consumerId) as NumOfCust, mm, yy FROM
(
    SELECT consumerId, month(date) as mm, year(date) as yy, sum(amount) as total, mdate FROM beta.transaction as t
    LEFT JOIN (
            SELECT min(month(date)) as mdate, consumerId as con FROM beta.transaction
            GROUP BY consumerId
            ) as MinDate ON con = t.consumerId
    GROUP BY month(date), consumerId
    HAVING mdate < mm AND total > 0
) as res
GROUP BY res.mm;

我会试着从里到外解释它

让我们看看名为 minDate 的 JOIN 表有什么：

SELECT min(month(date)) as mdate, consumerId as con FROM beta.transaction
GROUP BY consumerId

-- Here we find the first date of transaction per consumerId

下一个

SELECT consumerId, month(date) as mm, year(date) as yy, sum(amount), mdate FROM beta.transaction as t
    LEFT JOIN (
            SELECT min(month(date)) as mdate, consumerId as con FROM beta.transaction
            GROUP BY consumerId
            ) as MinDate ON con = t.consumerId
    GROUP BY month(date), consumerId
    HAVING mdate < mm AND total > 0

 -- Here we find total amount per consumerId per month and count only the consumers whose first transact (aka minDate) is lower than current month AND total is greater than 0

最后我们用外部 SELECT 来统计上面的按月分组的结果。我希望它是你想要的。

【讨论】：

非常感谢您的回复！我稍微更改了您的查询，并在下面发布了一个符合我需要的答案。

【解决方案2】：

下面的查询返回正确的结果。答案基于 Akis 的答案，即在 group by 子句中添加 consumerId 并在支票中添加年份和月份。

select
    tyyyy, tmm, count(tcon) as oldkund_real
from
(
    select
        t.yyyy as tyyyy, t.mm as tmm, t.consumerId as tcon, sum(t.amount) as total, fp.yyyy as fpyyyy, fp.mm as fpmm, fp.consumerId as fpcon
    from
        (
            select
                year(date) as yyyy, month(date) as mm, consumerId, amount
            from
                transaction
        ) as t
        left join
        (
            select
                year(min(date)) as yyyy, month(min(date)) as mm, consumerId
            from
                transaction
            group by
                consumerid
        ) as fp
        on
            t.consumerId = fp.consumerId
        group by
            t.yyyy, t.mm, t.consumerId
        having
            STR_TO_DATE(CONCAT('01,', tmm, ',', tyyyy),'%d,%m,%Y') > STR_TO_DATE(CONCAT('01,', fpmm, ',', fpyyyy),'%d,%m,%Y') and total > 0
) as res
group by
    tyyyy, tmm
order by
    tyyyy, tmm;

非常感谢阿基斯！

【讨论】：