Question

在使用JOINS一段时间后，我终于设法编写了一个没有语法错误的查询。但是，我需要根据日期限制它输出的数据。

简介是选择客户的ID，然后查看他们已兑换的忠诚度积分总数，并获得他们兑换的最后日期。我的查询可以收集客户ID，忠诚度会员ID和积分数，但是我得到他们兑换的每个日期，而不仅仅是最后一个。

有没有办法让查询只给出给定客户ID的最后日期？

我的查询是：

SELECT  gv.amount AS gv_amount, 
        gv.customer_id AS gv_customer_id, 
        h.date AS h_date, 
        c.loyalty_id AS c_loyalty_id
FROM coupon_gv_customer gv
   INNER JOIN loyalty_codes_redeem_history h 
        ON gv.customer_id = h.customer_id
    INNER JOIN customers c 
        ON gv.customer_id = c.customers_id

h_date的格式为YYYY-MM-DD HH：MM：SS

Answer 1

这并不难。请看一下我之前回答的这篇文章，了解使用JOIN的一些背景知识。

Selecting multiple columns/fields in MySQL subquery

这是您需要的技巧：一个子查询，用于查找每个客户ID的最新日期。这是：

 SELECT customer_id, MAX(date) as date
   FROM loyalty_codes_redeem_history
  GROUP BY customer_id

然后，您需要将此子查询（此虚拟表）加入到整个查询中。我认为结果将如下所示：

     SELECT gv.amount                 AS gv_amount, 
            gv.customer_id            AS gv_customer_id, 
            customer_latest_date.date AS h_date, 
            c.loyalty_id              AS c_loyalty_id
       FROM coupon_gv_customer gv
 INNER JOIN loyalty_codes_redeem_history h 
               ON gv.customer_id = h.customer_id
 INNER JOIN customers c 
               ON gv.customer_id = c.customers_id 
 INNER JOIN (
                 SELECT customer_id, 
                        MAX(date) AS date
                   FROM loyalty_codes_redeem_history
               GROUP BY customer_id
            ) customer_latest_date
               ON customer_latest_date.customer_id = c.customers_id

你看到它是如何工作的吗？子查询在INNER JOIN中使用，就好像它是一个表，实际上它是：一个虚拟表。

修改

要总结coupon_gv_customer表中的忠诚度积分，您需要另外一个摘要查询，如下所示：

SELECT customer_id, SUM(amount) AS amount FROM coupon_gv_customer GROUP BY customer_id

然后，我认为您的查询将是这样。每个客户会给你一排，我认为这是你想要的。没有任何兑换或日期的客户将不会出现。

SELECT c.customers_id, c.loyalty_id, customer_latest_date.date customer_points.amount FROM customers c INNER JOIN ( SELECT customer_id, SUM(amount) AS amount FROM coupon_gv_customer GROUP BY customer_id ) customer_points ON c.customers_id = customers_points.customer_id INNER JOIN ( SELECT customer_id, MAX(date) AS date FROM loyalty_codes_redeem_history GROUP BY customer_id ) customer_latest_date ON customer_latest_date.customer_id = c.customers_id

将sql查询的结果限制为按客户ID输入的最后日期

1 个答案: