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
   )
)

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
   )
)

我需要你需要的东西，尽管是在许多年后，我尝试了最流行的两个答案。这些都没有产生预期的结果。这就是我所能提供的…为了清晰起见，我更改了一些名称。

SELECT 
  cc.pk_ID AS pk_Customer_ID, 
  cc.Customer_Name AS Customer_Name, 
  IFNULL(pp.pk_ID, '') AS fk_Purchase_ID,
  IFNULL(pp.fk_Customer_ID, '') AS fk_Customer_ID,
  IFNULL(pp.fk_Item_ID, '') AS fk_Item_ID,
  IFNULL(pp.Purchase_Date, '') AS Purchase_Date
FROM customer cc
LEFT JOIN purchase pp ON (
  SELECT zz.pk_ID 
  FROM purchase zz 
  WHERE cc.pk_ID = zz.fk_Customer_ID 
  ORDER BY zz.Purchase_Date DESC LIMIT 1) = pp.pk_ID
ORDER BY cc.pk_ID;

表:

Customer => id, name
Purchase => id, customer_id, item_id, date

查询:

SELECT C.id, C.name, P.id, P.date
  FROM customer AS C
  LEFT JOIN purchase AS P ON 
    (
      P.customer_id = C.id 
      AND P.id IN (
        SELECT MAX(PP.id) FROM purchase AS PP GROUP BY PP.customer_id
      )
    )

你也可以指定一些条件到子选择查询

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。

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;