Question

我正在尝试获取用户的排名的二维参数：donation sum和total donor count。

我的等级公式为：rank of [rank of donation_sum + rank of donor_count / 2]

样本表：

donation_id | user_id | donor_id | donation_sum
-----------------------------------------------
1 | 1 | 1 | 10
2 | 1 | 2 | 5
3 | 2 | 3 | 10
4 | 3 | 1 | 50
...

如您所见，有些捐赠者向不同的用户捐款，因此我使用sum(donation_sum)和count(distinct(donation_id))来获得准确的排名

我能够通过捐赠金额和捐赠总数分别获得2个sql的排名列表，但我需要的是在postgreSQL v.9.4中给定user_id以上公式的用户等级

你有任何解决方案吗？所以我将在Yii2 PHP框架中使用该sql查询

由于

编辑：我们将donation_date添加到tbl_donation并修改了实际查询，如下所示：

是否真实使用了donation_date？

    with list as (
        select 
        s.runner_id, sum, count, rank_sum, rank_count,
        (rank_sum+ rank_count)::float/ 2 as rank_avg,
        row_number() over (order by rank_sum) as rank
    from (
        select *, rank() over (order by sum desc) rank_sum
        from (
            select runner_id, sum(donation_sum)
            from tbl_donation
where donation_date >= '2015-01-01'
            group by 1
            ) s
        ) s
    join (  
        select *, rank() over (order by count desc) rank_count
        from (
            select runner_id, count(distinct(donator_id))
            from tbl_donation
where donation_date >= '2015-01-01'
            group by 1
            ) c
        ) c
    using (runner_id)
    )
    select rank
    from list
    where runner_id = 251;

Answer 1

在不同的子查询中进行两次排名：

select 
    s.user_id, sum, count, rank_sum, rank_count,
    (rank_sum+ rank_count)::float/ 2 as rank_avg,
    row_number() over (order by rank_sum) as rank
from (
    select *, rank() over (order by sum desc) rank_sum
    from (
        select user_id, sum(donation_sum)
        from donations
        group by 1
        ) s
    ) s
join (  
    select *, rank() over (order by count desc) rank_count
    from (
        select user_id, count(distinct(donation_id))
        from donations
        group by 1
        ) c
    ) c
using (user_id);

 user_id | sum | count | rank_sum | rank_count | rank_avg | rank 
---------+-----+-------+----------+------------+----------+------
       3 | 100 |     1 |        1 |          2 |      1.5 |    1
       1 |  30 |     2 |        2 |          1 |      1.5 |    2
       2 |  20 |     1 |        3 |          2 |      2.5 |    3
(3 rows)

如果您想为单rank次使用user_id选择with query，例如：

with list as (
    -- place here the above query
)
select rank
from list
where user_id = 2;

给定user_id

是否真实使用了donation_date？

1 个答案: