Question

我发现了一些示例，说明如何从分组集中选择单个最旧/最新的行，但是从数据集中获取最旧的两行时遇到了问题。

这是我的示例表：

CREATE TABLE IF NOT EXISTS `orderTable` (
  `customer_id` varchar(10) NOT NULL,
  `order_id` varchar(4) NOT NULL,
  `date_added` date NOT NULL,
  PRIMARY KEY (`customer_id`,`order_id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

INSERT INTO `orderTable` (`customer_id`, `order_id`, `date_added`) VALUES
('1234', '5A', '1997-01-22'),
('1234', '88B', '1992-05-09'),
('0487', 'F9', '2002-01-23'),
('5799', 'A12F', '2007-01-23'),
('1234', '3A', '2009-01-22'),
('3333', '7FHS', '2009-01-22'),
('0487', 'Z33', '2004-06-23'),
('3333', 'FF44', '2013-09-11'),
('3333', '44f5', '2013-09-02');

此查询返回两行以上：

SELECT customer_id, order_id, date_added
FROM orderTable T1
WHERE (
   select count(*) FROM orderTable T2
   where T2.order_id = T1.order_id AND T2.date_added <= T1.date_added
) <= 2;

由于我不是在寻找单行，因此这不是标准的greatest-n-per-group类型查询。

我错过了什么，我可以获得每个customer_id的前两个订单？

Answer 1

最佳（即最佳性能）方法是在查询中使用用户定义变量。

SELECT tmp.customer_id, tmp.date_added 
FROM ( 
  SELECT 
    customer_id, date_added, 
    IF (@prev <> customer_id, @rownum := 1, @rownum := @rownum+1 ) rank,
    @prev := customer_id
  FROM orderTable t 
  JOIN (SELECT @rownum := NULL, @prev := 0) r 
  ORDER BY t.customer_id
) tmp 
WHERE tmp.rank <= 2 
ORDER BY customer_id, date_added

<强>结果：

| CUSTOMER_ID |                       DATE_ADDED |
|-------------|----------------------------------|
|        0487 |   January, 23 2002 00:00:00+0000 |
|        0487 |      June, 23 2004 00:00:00+0000 |
|        1234 |       May, 09 1992 00:00:00+0000 |
|        1234 |   January, 22 1997 00:00:00+0000 |
|        3333 |   January, 22 2009 00:00:00+0000 |
|        3333 | September, 02 2013 00:00:00+0000 |
|        5799 |   January, 23 2007 00:00:00+0000 |

小提琴here。

请注意，连接仅用于初始化变量。

Answer 2

您的原始查询应该是（在子查询中使用customer_id）

SELECT customer_id, order_id, date_added
FROM orderTable T1
WHERE (
   select count(*) FROM orderTable T2
   where T2.customer_id = T1.customer_id AND T2.date_added <= T1.date_added
) <= 2;

您也可以使用变量：

SELECT customer_id, order_id, date_added FROM (
SELECT customer_id, order_id, date_added,
@rownum := if(@prev_cust = customer_id, @rownum + 1,1) as rn,
@prev_cust := customer_id cust_var
FROM orderTable T1,
  (SELECT @rownum := 0) r,
  (SELECT @prev_cust := '') c
order by customer_id, date_added 
) o where o.rn < 3;

SQL DEMO

Answer 3

这是另一种（故意不完整的）方法，尽管其他人可能对性能有所了解......

SELECT x.*
     , COUNT(*) rank 
  FROM ordertable x 
  JOIN ordertable y 
    ON y.customer_id = x.customer_id 
   AND y.date_added <= x.date_added 
 GROUP 
    BY x.customer_id 
     , x.date_added;

Answer 4

这应该产生你所追求的结果，但外部SELECT不会是最有效的，因为它在派生表上进行过滤。

SELECT ranked.* 
FROM (
    SELECT ot.* , 
        @rownum := IF( ot.customer_id = @previous , @rownum +1, 1 ) rank, 
        @previous := ot.customer_id
    FROM orderTable ot, 
        (SELECT @rownum :=1, @previous := NULL) init
    ORDER BY customer_id, date_added
) ranked
WHERE rank <=2

从组中选择最旧的两个记录

4 个答案: