获得最高数量的产品/用户数量

时间:2017-10-30 17:58:52

标签: php mysql group-by sum highest

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

我想从上表中得到2个结果:

首先by 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

第二个by user_id如下:

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:上方: user_id = 2已对产品10,20,30,40(对prd_id出价)出价,因此他的出价续= 4 ...而且他的出价是获胜者10,40(竞价prod_id的赢家)..WHY仅10,40而不是30 ... bcz user_id = 4对prd = 30的出价,amt = 100,user_id = 2,amt = 100 ..但是出价是由用户= 4在prd = 30上进行的,因此他是prd = 30的赢家(同样的amt)

尝试by prd_id的查询,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`

以上查询结果如下:( user_id,disc不合适)

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by user_id,我无法获得将要查询的内容。

由于

更新:

感谢HARSHIL:http://www.sqlfiddle.com/#!9/5325a6/5/1

但是经过一些更多的输入我发现了这个错误:http://www.sqlfiddle.com/#!9/e04063/1代表第二个:代表user_id 但适用于 prd_id(首次查询)

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null 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

使用null user_id (但在我的wamp服务器中,我在win_on_bid_prd_id中看不到user_id = 1的空值,我得到值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

1 个答案:

答案 0 :(得分:4)

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;

Click here for UPDATED DEMO