php-获得最高数量,产品/用户数
作者:互联网
id | user_id | prd_id | amnt | dis
1 | 1 | 10 | 200 | 23
2 | 2 | 10 | 300 | 11
3 | 3 | 20 | 100 | 26
4 | 2 | 20 | 50 | 12
5 | 4 | 30 | 100 | 22
6 | 2 | 40 | 600 | 18
7 | 2 | 30 | 100 | 16
我想要上表2个结果:
首先通过prod_id如下
prd_id | user_id | cont | highestamt | disc
10 | 2 | 2 | 300 | 11
20 | 3 | 2 | 100 | 26
30 | 4 | 2 | 100 | 22
40 | 2 | 1 | 600 | 18
其次是user_id,如下所示:
user_id | cont | bid on prd_id | winner on bid prod_id |
1 | 1 | 10 | - | -
2 | 4 | 10,20,30,40 | 10,40 |
3 | 1 | 20 | 20 |
4 | 1 | 30 | 30 |
更新:例如:上面:user_id = 2对产品10,20,30,40的出价(对prd_id的出价),因此他的出价cont = 4 …而他在10,40中的中标(prod_id的中标) )..为什么只有10,40而不是30 … bcz user_id = 4对ard = 100的prd = 30出价,而对amt = 100 ..的user_id = 2出价,但是user = 4对prd =出价30,因此他是prd = 30的赢家(相同的amt)
尝试通过prd_id查询以下内容,但它给了我一些错误的结果.
SELECT `prd_id`, `user_id` , count('prd_id') as cont , max(`amnt`) as highestamt,disc
FROM `proddtails`
group by `prd_id` order by `prd_id`
上面的查询结果如下:(user_id,disc不正确)
prd_id | user_id | cont | highestamt | disc
10 | 2 | 2 | 300 | 11
20 | 2 | 2 | 100 | 11
30 | 2 | 1 | 100 | 11
40 | 2 | 1 | 600 | 11
对于第二个user_id,我没有得到什么查询.
谢谢
更新:
感谢HARSHIL:http://www.sqlfiddle.com/#!9/5325a6/5/1
但经过更多输入后,我发现此错误:第二个http://www.sqlfiddle.com/#!9/e04063/1:对于user_id,但对于prd_id效果很好(第一个查询)
user_id cont bid_on_prd_id winner_on_bid_prod_id
1 1 10 (null)
2 4 10,20,40,30 10,40,30
3 1 20 20
4 1 30 30
但我想如下:
没有空的user_id
user_id cont bid_on_prd_id winner_on_bid_prod_id
2 4 10,20,30,40 10,40
3 1 20 20
4 1 30 30
user_id为null(但在我的Wamp服务器中,在user_id = 1的winner_on_bid_prd_id中我看不到null,我得到的值是10而不是null)
user_id cont bid_on_prd_id winner_on_bid_prod_id
1 1 10 (null)
2 4 10,20,30,40 10,40
3 1 20 20
4 1 30 30
解决方法:
对于prd_id:
select t1.prd_id,t1.user_id,
(select count(*) from tablename where prd_id = t1.prd_id)as cont,
t1.amnt as highststatment,
t1.dis as disc
from tablename t1
where (t1.prd_id,t1.amnt) in
(select prd_id, max(amnt) from tablename group by prd_id)
group by t1.prd_id;
对于usr_id:
select t1.user_id,
count(*) as cont,
Group_concat(t1.prd_id separator ',') as bid_on_prd_id,
(select Group_concat(distinct t2.prd_id separator ',')
from tablename t2
where t2.user_id = t1.user_id
and (t2.id) in
(select min(id) from tablename
where (prd_id,amnt) in
(select prd_id,max(amnt) from tablename group by prd_id)
group by prd_id
)
) as winner_on_bid_prod_id
from tablename t1
group by t1.user_id
having winner_on_bid_prod_id IS NOT NULL;
标签:sum,mysql,php,group-by,highest 来源: https://codeday.me/bug/20191025/1929754.html