MYSQL - 获取用户和最近的购买数据答案

【问题标题】：MYSQL - Get User and Most Recent Purchase DataMYSQL - 获取用户和最近的购买数据
【发布时间】：2017-04-04 06:39:58
【问题描述】：

我正在尝试执行 MYSQL 查询，在其中获取用户最近的购买记录，然后查看是否符合特定条件。这是我放在一起的查询：

select
    users_purch.purch_date as purchase_date,
    users_purch.total_amount as purchase_amount,
    users.* 
from
    users 
left join
    (
        select
            max(date) as purch_date,
            user_id,
            total_amount 
        from
            users_purchases 
        group by
            user_id
    ) as users_purch 
        on users_purch.user_id = users.id 
where
    users_purch.purch_date < '2016-11-01' 
    and users_purch.total_cost < 112.49 
order by
    users_purch.purch_date desc

查询似乎有效，但在某些方面失败了。例如，如果用户有多个购买条目，则它会获取最大日期，但查询检索到的 total_cost 金额与最大日期不在同一行。如何重写此查询以提供最新的完整购买记录？

谢谢！

【问题讨论】：

发布您的表格概览会有所帮助。我们可以根据表格布局（也许）为您提供一些关于查询的新见解
您的内部查询 users_purch 似乎是错误的。是total_amount 是sum(amount)??您的 Group by 只有一列，但您在 select 中有 2 列以及一个聚合函数
这和php标签有什么关系？
请注意，LEFT JOIN 在此处作为常规 INNER JOIN 执行，这要归功于 WHERE 子句条件。
感谢大家的cmets。很有帮助！

标签： php mysql sql database

【解决方案1】：

您必须再次加入user_purchases 表才能获取有关日期的信息：

select
    users_purch.purch_date as purchase_date,
    users_purch.total_amount as purchase_amount,
    users.* 
from
    users 
left join
    (
        select
            max(date) as purch_date,
            user_id
        from
            users_purchases 
        group by
            user_id
    ) as users_purch 
        on users_purch.user_id = users.id 
left join
   (
       select
          user_id,
          date,
          total_amount
       from
          users_purchases
   ) as users_purch2 on users_purch.user_id = users_purch2.user_id and
        users_purch2.date = users_purch.purch_date
where
    users_purch.purch_date < '2016-11-01' 
    and users_purch.total_cost < 112.49 
order by
    users_purch.purch_date desc

【讨论】：

非常感谢。这工作得很好。我知道我必须重新加入那张桌子，但我一生都无法弄清楚如何去做。看了这个之后，我想知道从“users_purch”中获取ID并在与“users_purch2”的连接中使用它是否是一个更强大的连接？
@ackerchez 如果您尝试获取ID 值，那么您基本上会遇到同样的问题：获得的ID 值不能保证是那个值对应于最新的日期值。
我明白了。所以只要我没有在同一秒购买两次（似乎不可能）并且由同一个用户购买，那么我应该没问题。
@ackerchez 是的，在这种情况下，关键是用户、日期值对。

【解决方案2】：

由于您在此选择中的分组

如果您需要最大（日期）的数量，则重写以提取适当的数量

  select
      users_purch.purch_date as purchase_date,
      users_purch.total_amount as purchase_amount,
      users.* 
  from
      users 
  left join
      (
        select t1.purch_date, t1.user_id, t2.total_amount from (
         select
                max(date) as purch_date,
                user_id
          from  users_purchases 
            group by  user_id ) t1
            inner join (
              date,
              user_id,
              total_amount 
          from users_purchases 
              ) t2 on t1.user_id= t2.user_id, t1.purch_date = t2.date

      ) as users_purch 
          on users_purch.user_id = users.id 
  where
      users_purch.purch_date < '2016-11-01' 
      and users_purch.total_cost < 112.49 
  order by
      users_purch.purch_date desc

【讨论】：

【解决方案3】：

你可以试试这个，伙计：

SELECT
    up.date AS 'purchase_date', 
    up.total_amount AS 'purchase_amount',
    u.*
FROM 
    users u
    INNER JOIN users_purchases up ON up.user_id = u.user_id
    INNER JOIN (
        # get max date per user_id
        SELECT user_id, max(date) AS 'purch_date'
        FROM users_purchases 
        GROUP BY user_id
    ) max_up ON 
        max_up.user_id
        # join that date to get the correct total_amount
        AND max_up.`purch_date` = up.`date`
WHERE 
    up.`date` < '2016-11-01'
    AND up.total_amount < 112.49
GROUP BY u.user_id
ORDER BY up.`date` DESC;

注意

user_id 的最大日期
total_amount 每个最大日期的值
按user_id 分组
使用users_purchases.date 降序排列

【讨论】：