您还没有指定数据库。如果它是一个允许分析函数的方法，那么使用这种方法可能比GROUP BY更快(在Oracle中肯定更快，在SQL Server的后期版本中很可能更快，不知道其他版本)。

SQL Server中的语法是:

SELECT c.*, p.*
FROM customer c INNER JOIN 
     (SELECT RANK() OVER (PARTITION BY customer_id ORDER BY date DESC) r, *
             FROM purchase) p
ON (c.id = p.customer_id)
WHERE p.r = 1

先不讲代码，逻辑/算法如下:

Go to the transaction table with multiple records for the same client. Select records of clientID and the latestDate of client's activity using group by clientID and max(transactionDate) select clientID, max(transactionDate) as latestDate from transaction group by clientID inner join the transaction table with the outcome from Step 2, then you will have the full records of the transaction table with only each client's latest record. select * from transaction t inner join ( select clientID, max(transactionDate) as latestDate from transaction group by clientID) d on t.clientID = d.clientID and t.transactionDate = d.latestDate) You can use the result from step 3 to join any table you want to get different results.

请尝尝这个，

SELECT 
c.Id,
c.name,
(SELECT pi.price FROM purchase pi WHERE pi.Id = MAX(p.Id)) AS [LastPurchasePrice]
FROM customer c INNER JOIN purchase p 
ON c.Id = p.customerId 
GROUP BY c.Id,c.name;

这是经常出现在StackOverflow上的“每组最多n个”问题的一个例子。

以下是我通常建议的解决方法:

SELECT c.*, p1.*
FROM customer c
JOIN purchase p1 ON (c.id = p1.customer_id)
LEFT OUTER JOIN purchase p2 ON (c.id = p2.customer_id AND 
    (p1.date < p2.date OR (p1.date = p2.date AND p1.id < p2.id)))
WHERE p2.id IS NULL;

解释:给定第p1行，不应该有第p2行具有相同的客户和较晚的日期(或者在领带的情况下，较晚的id)。当我们发现这是正确的，那么p1是该客户最近的购买。

关于索引，我将在purchase列(customer_id、date、id)上创建一个复合索引。这可能允许使用覆盖索引来完成外部连接。请确保在您的平台上进行测试，因为优化是依赖于实现的。使用RDBMS的特性来分析优化计划。例如，MySQL上的EXPLAIN。

有些人使用子查询而不是我上面展示的解决方案，但我发现我的解决方案更容易解决关系。

另一种方法是在你的连接条件中使用NOT EXISTS条件来测试以后的购买:

SELECT *
FROM customer c
LEFT JOIN purchase p ON (
       c.id = p.customer_id
   AND NOT EXISTS (
     SELECT 1 FROM purchase p1
     WHERE p1.customer_id = c.id
     AND p1.id > p.id
   )
)

如果你正在使用PostgreSQL，你可以使用DISTINCT ON来查找组中的第一行。

SELECT customer.*, purchase.*
FROM customer
JOIN (
   SELECT DISTINCT ON (customer_id) *
   FROM purchase
   ORDER BY customer_id, date DESC
) purchase ON purchase.customer_id = customer.id

PostgreSQL Docs - Distinct On

注意，DISTINCT ON字段——这里是customer_id——必须匹配ORDER BY子句中最左边的字段。

注意:这是一个非标准条款。