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

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;

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]

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
   )
)

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.

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