Question

我有一个SQL表包含销售某些项目的数据。事实上，它有物品销售的日志。

例如，有一个包含2个项目的促销：键盘（id:1）和鼠标（id:2）。买家可以多次对每件商品进行出价，比如ebay。因此，我们假设有2位买家（ids are 97 and 98）多次出价。相关数据将是：

bid_id  |   buyer_id    |   item_id |   amount      |   time                |
1       |   97          |   1       |   44.26       |   2014-01-20 15:53:16 |
2       |   98          |   2       |   30.47       |   2014-01-20 15:54:52 |
3       |   97          |   2       |   40.05       |   2014-01-20 15:57:47 |
4       |   97          |   1       |   42.46       |   2014-01-20 15:58:36 |
5       |   97          |   1       |   39.99       |   2014-01-20 16:01:13 |
6       |   97          |   2       |   24.68       |   2014-01-20 16:05:35 |
7       |   98          |   2       |   28          |   2014-01-20 16:08:42 |
8       |   98          |   2       |   26.75       |   2014-01-20 16:13:23 |

在此表中，我需要为每个用户选择第一项商品的数据，并为每个用户选择最后的商品。

因此，如果我为每个用户选择第一项商品（不同），则返回数据应为：

bid_id  |   buyer_id    |   item_id |   amount      |   time                |
1       |   97          |   1       |   44.26       |   2014-01-20 15:53:16 |
2       |   98          |   2       |   30.47       |   2014-01-20 15:54:52 |
3       |   97          |   2       |   40.05       |   2014-01-20 15:57:47 |

如果我为每个用户选择最后一个优惠，则返回应该是：

bid_id  |   buyer_id    |   item_id |   amount      |   time                |
5       |   97          |   1       |   39.99       |   2014-01-20 16:01:13 |
6       |   97          |   2       |   24.68       |   2014-01-20 16:05:35 |
8       |   98          |   2       |   26.75       |   2014-01-20 16:13:23 |

由于我必须为每个用户提供每个项目，因此我尝试GROUP BY和buyer_id item_id，然后SELECT MIN的{{1}}值time或bid_id。但它始终首先返回bid_id但最新的amount行（实际上是最后一次提供）。

这是我试过的查询：

SELECT MIN(`bid_id`) AS `bid_id`,`buyer_id`,`item_id`,`amount`,`time` FROM `offers` GROUP BY `buyer_id`,`item_id`

结果是：

bid_id  |   buyer_id    |   item_id |   amount      |   time                |
1       |   97          |   1       |   39.99       |   2014-01-20 16:01:13 |
2       |   97          |   2       |   24.68       |   2014-01-20 16:05:35 |
3       |   98          |   2       |   26.75       |   2014-01-20 16:13:23 |

如您所见，它分组依据且ID正确但其余行值不正确。

分组购买多列时，如何正确SELECT第一行和/或最后一行？

Answer 1

SELECT o.`bid_id`,o.`buyer_id`,o.`item_id`,o.`amount`,o.`time` FROM `offers` o
JOIN
(SELECT MIN(`bid_id`) AS `bid_id`,`buyer_id`,`item_id`,`amount`,`time` FROM `offers` GROUP BY `buyer_id`,`item_id`)x
ON x.bid_id=o.bid_id  AND x.buyer_id=o.buyer_id

Answer 2

这是另一种观点，使用Quassnoi's ranking trick here

首次出价：

SELECT x.bid_id, x.buyer_id, x.item_id, x.amount, x.time 
FROM 
(
  SELECT o.bid_id, o.buyer_id, o.item_id, o.amount, o.time, 
    @combo :=CASE WHEN NOT(@curItem = o.item_id AND @curBuyer = o.buyer_id) 
                  THEN 1 ELSE @combo+1 END AS Rank,
    @curItem:=o.item_id AS item,
    @curBuyer:=o.buyer_id AS buyer
  FROM
  (
    SELECT o.bid_id, o.buyer_id, o.item_id, o.amount, o.time 
      FROM offers o
      ORDER BY o.buyer_id, o.item_id, o.bid_id
  ) o,
  (SELECT @curItem := -1) itm,
  (SELECT @curBuyer:= -1) buy
) x
WHERE x.Rank = 1;

对于上次出价查询，您只需将ORDER BY更改为o.buyer_id, o.item_id, o.bid_id DESC

SqlFiddle here

Answer 3

首先提供sql：

SELECT
*
FROM
    offers AS o1
WHERE
NOT EXISTS (
    SELECT
        1
    FROM
        offers o2
    WHERE
        o1.buyer_id = o2.buyer_id
    AND o1.item_id = o2.item_id
    AND datetime(o1.time) > datetime(o2.time)
)

last offer sql：只需更改为datetime（o1.time）＆lt; datetime（o2.time）（我使用sqlite~）

Answer 4

请在下面查询所需的输出。 SQL FIDDLE LINK：http://sqlfiddle.com/#!2/916c2/15

    (select f.bid_id,f.buyer_id,f.item_id,f.amount,f.time from offers f join
    (select buyer_id,item_id,min(time) as time from offers
    group by buyer_id,item_id)t
    on f.buyer_id=t.buyer_id and f.item_id=t.item_id
    and f.time=t.time)
    union
    (select f.bid_id,f.buyer_id,f.item_id,f.amount,f.time from offers f join
    (select buyer_id,item_id,max(time) as time from offers
    group by buyer_id,item_id)t
    on f.buyer_id=t.buyer_id and f.item_id=t.item_id
    and f.time=t.time);

MySQL：按多列分组时选择第一行或最后一行

4 个答案: