[英]MySQL: SELECT First or Last Rows When Grouping By Multiple Columns
我有一个SQL表,其中包含一些项目的销售数据。 实际上,它具有项目销售的日志。
例如,某笔交易包含2个项目:键盘( id:1
)和鼠标( id:2
)。 买家可以对每个项目进行多次出价,例如ebay。 因此,假设有2个买家( ids are 97 and 98
)进行了几次出价。 相关数据为:
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 |
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,我需要为每个用户的第一个商品报价和每个用户的最后一个报价选择数据。
因此,如果我为每个用户(不同)选择第一项报价,则返回数据应类似于:
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,则退货应为:
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 |
由于必须为每个用户携带每件商品,因此我尝试对buyer_id
和item_id
都buyer_id
GROUP BY
,然后SELECT
time
或bid_id
的MIN
值。 但是,它总是向我返回第一个bid_id
但返回最近的amount
行(实际上是最后报价)。
这是我尝试过的查询:
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,它的分组依据和ID是正确的,但其余的行值却不是。
分组购买多列时如何正确SELECT
第一行和/或最后一行?
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
这是另一个使用Quassnoi的排名技巧的方法
对于第一个出价:
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;
对于最后一个出价查询,您只需将ORDER BY
更改为o.buyer_id, o.item_id, o.bid_id DESC
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)
)
最后提供sql:只需更改为datetime(o1.time)<datetime(o2.time)(我使用sqlite〜)
请尝试在下面的查询中查询所需的输出。 SQL链接: 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);
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