Mysql 我可以在分组之前执行COUNT（）吗

我正在为学校做一个mySQL作业，我被一个问题困住了。我还是mySQL的新手。计数（o.customer_id）没有按我想要的方式工作。我想让它计算订单数量，但它正在计算所有项目。i、 e.客户1有2个订单，但它返回3个，因为一个订单有两个项目。我有三张桌子，一张是顾客桌，另一张是订单桌，每张订单上的每一项都有。我在下面提出了我的问题。任何帮助都会很好

SELECT email_address, COUNT(o.order_id) AS num_of_orders,
       SUM(((item_price - discount_amount) * quantity)) AS total
FROM customers c JOIN orders o
      ON c.customer_id = o.customer_id
     JOIN order_items ot
      ON o.order_id = ot.order_id
GROUP BY o.customer_id
HAVING num_of_orders > 1
ORDER BY total DESC;

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看起来您要计算不同的订单数。在计数中添加一个不同的值。尽管MySQL允许您在HAVING子句中使用SELECT表达式，但这样做并不是一种好的做法

SELECT email_address, COUNT(DISTINCT o.order_id) AS num_of_orders,
       SUM(((item_price - discount_amount) * quantity)) AS total
FROM customers c JOIN orders o
      ON c.customer_id = o.customer_id
     JOIN order_items ot
      ON o.order_id = ot.order_id
GROUP BY o.customer_id
HAVING COUNT(DISTINCT o.order_id) > 1
ORDER BY total DESC;

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只需使用

不同的
保留字即可：

SELECT email_address, COUNT(distinct o.order_id) AS num_of_orders

只需取出项目的连接。它所做的只是在有多个项目时复制行

SELECT email_address, COUNT(o.order_id) AS num_of_orders,
       SUM(((item_price - discount_amount) * quantity)) AS total
FROM customers c JOIN orders o
      ON c.customer_id = o.customer_id
GROUP BY o.customer_id
HAVING COUNT(o.order_id) > 1
ORDER BY total DESC;

改变一行对我来说也会产生同样的结果！不适用于mysql。。。（参见第二个选择示例）