MySQL - 选择在过去 y 天内至少下了 x 个订单并且在过去 z 天内没有订购的客户

Question

我正在尝试获取在过去 y 天内至少下了 x 订单并且在过去 z 内没有下订单的客户code> 天，所以我的客户可以知道哪些客户正在“离开”并且不会回来。

我的表很简单:

订单

- ID
- Created (UNIX timestamp)
- CustomerID

客户

- ID
- Name
- Phone

我在过去的 y 天内成功选择了 x 个订单，但我无法加入数据，所以我将只获得未订购的客户最近 z 天

这是我对在过去 30 天内下了 1 个以上订单的客户的查询:

SELECT
    COUNT(t1.CustomerID) as customer_count,
    t2.ID as customer_id,
    t2.Name,
    t2.Phone
FROM
    Orders t1
JOIN
    Customers t2
        ON t1.CustomerID = t2.ID
WHERE
    t1.Created BETWEEN (UNIX_TIMESTAMP() - (86400*30)) AND UNIX_TIMESTAMP()
GROUP BY
    t2.ID
HAVING customer_count >= 2

Answer 1

由于您在第一部分中取得了成功，而第二部分基本相同，您可以通过复制查询、稍作修改并将其作为子查询添加到 HAVING 来简化操作健康)状况。对过去 7 天内未下任何订单的客户试试这个:

SELECT
    COUNT(o1.CustomerID) as customer_count,
    c1.ID as customer_id,
    c1.Name,
    c1.Phone
FROM Customers c1
LEFT JOIN Orders o1 ON o1.CustomerID = c1.ID
WHERE o1.Created BETWEEN (UNIX_TIMESTAMP() - (86400*30)) AND UNIX_TIMESTAMP()
GROUP BY c1.ID
HAVING customer_count > 1
   AND (SELECT 
            COUNT(o2.CustomerID) as customer_count,
        FROM Customers c2
        LEFT JOIN Orders o2 ON o2.CustomerID = c2.ID
        WHERE o2.Created BETWEEN (UNIX_TIMESTAMP() - (86400*7)) AND UNIX_TIMESTAMP()
          AND c2.ID = c1.ID
        GROUP BY c2.ID) = 0

注意事项:

• 我将您的条件从 customer_count >= 2 更改为 customer_count > 1 因为它读起来更好。
• 我建议从主表 Customers 开始，然后加入 Orders，因为这样感觉更自然。
• 我还建议使用比 t1 和 t2 更有意义的别名。例如，c 和 o 读起来更好。通过上面的查询，您可以执行类似 c1、c2、o1 和 o2 的操作。

编辑 由于子查询只计算一个给定客户的订单，我们不需要分组所以我们可以像这样简化它:

SELECT
    COUNT(o.CustomerID) as customer_count,
    c.ID as customer_id,
    c.Name,
    c.Phone
FROM Customers c
LEFT JOIN Orders o ON o.CustomerID = c.ID
WHERE o.Created BETWEEN (UNIX_TIMESTAMP() - (86400*30)) AND UNIX_TIMESTAMP()
GROUP BY c.ID
HAVING customer_count > 1
   AND (SELECT COUNT(*)
        FROM Orders o2
        WHERE o2.Created BETWEEN (UNIX_TIMESTAMP() - (86400*7)) AND UNIX_TIMESTAMP()
          AND o2.CustomerID = c.ID) = 0