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;

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
   )
)

试试这个，会有帮助的。

我在我的项目中使用了这个。

SELECT 
*
FROM
customer c
OUTER APPLY(SELECT top 1 * FROM purchase pi 
WHERE pi.customer_id = c.Id order by pi.Id desc) AS [LastPurchasePrice]

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

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;

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。