2

假设，我有一个包含以下列的表格：

date | event     | user_id |  unit_id  |cost |  ad_id  | spend
03-15 | impression | 2353   |   3436 | 0.15 |  NULL | NULL
03-15 | impression | 2353   |   3436 | 0.12 |  NULL | NULL
03-15 | impression | 1234   |   5678 | 0.10 |  NULL | NULL
03-15 | click | 1234   |   5678 |  NULL  |  NULL | NULL
03-15 | create_ad |     1234 | 5678 | NULL | 6789 | 10

我想计算在用户创建 ID 之前平均需要多少次展示。在此特定场景中，用户 1234 需要一次印象来创建广告。

我不确定我能否以某种方式使用日期来区分事件（但从逻辑上讲，所有这些事件都应该在不同的时刻发生）。但是，您可以看到展示次数在ad_id和支出中具有 NULL，而 create_id 在支出中确实有一个数字。

这个不行：

select i.user_id
     , i.unit_id
     , count(i.event) impressions_n
     , count(c.event) as ads_n
  from add4ad i
   left 
  join add4ad c
     on i.user_id = c.user_id 
   and i.unit_id = c.unit_id
 where i.event in ('impression')
   and c.spend <> NULL
 group 
    by i.user_id
     , i.unit_id

我用这些数据创建了一个SQLFiddle

4

4 回答 4

2

我去了 SQL Fiddle 并通过 MS SQL 引擎运行了测试。

CREATE TABLE add4ad (date date, event varchar(10), user_id int,
                   unit_id int, cost float, ad_id float, spend float);
INSERT INTO add4ad (date, Event, user_id,unit_id,cost,ad_id,spend)
VALUES
    ('2018-03-15','impression','2353','3436','0.15',NULL,NULL),
    ('2018-03-15','impression','2353','3436','0.12',NULL,NULL),
    ('2018-03-15','impression','2353','3436','0.10',NULL,NULL),
    ('2018-03-15','click','1234','5678', NULL, NULL,NULL),
    ('2018-03-15','create_ad','2353','5678', NULL, 6789,10);

我的查询

with e10 as (select  user_id, event, date, rowid=row_number() over (Partition by user_id order by date)
from add4ad
where event='create_ad'
),
e20 as ( -- get the first create_ad event
select user_id, date
  from e10
  where rowid=1
  )
  select a.user_id, count(1) as N
  from e20 inner join add4ad a
  on e20.user_id=a.user_id
  and a.date<=e20.date
  and a.event='impression'
  group by a.user_id
于 2019-12-26T20:37:41.097 回答
2

如果我猜对了，您需要计算不同的广告

CREATE TABLE add4ad (`date` date, `event` varchar(10), `user_id` int,
                   `unit_id` int, `cost` float, `ad_id` float, `spend` float);
INSERT INTO add4ad (`date`, `Event`, `user_id`,`unit_id`,`cost`,`ad_id`,`spend`)
VALUES
    ('2018-03-15','impression','2353','3436','0.15',NULL,NULL),
    ('2018-03-15','impression','2353','3436','0.12',NULL,NULL),
    ('2018-03-15','impression','2353','3436','0.10',NULL,NULL),
    ('2018-03-15','impression','1234','5678','0.10',NULL,NULL),
    ('2018-03-15','click','1234','5678', NULL, NULL,NULL),
    ('2018-03-15','create_ad','1234','5678', NULL, 6789,10),
    ('2018-03-16','impression','8765','8871','0.10',NULL,NULL),
    ('2018-03-16','impression','8765','8871','0.10',NULL,NULL),
    ('2018-03-16','impression','8765','8871','0.2',NULL,NULL),
    ('2018-03-16','impression','8765','8871','0.23',NULL,NULL),
    ('2018-03-16','click','8765','8871', NULL, NULL,NULL),
    ('2018-03-16','create_ad','8765','8871', NULL, 6789,10);

select i.user_id, i.unit_id, count(i.event) as impressions_n,
    count(distinct c.event) as ads_n
from add4ad i
join add4ad c
   on i.user_id = c.user_id and i.unit_id = c.unit_id
where i.event in ('impression')
   and c.event in ('create_ad') and c.spend is not NULL
group by i.user_id, i.unit_id

退货

user_id unit_id impressions_n   ads_n
1234    5678    1   1
8765    8871    4   1

我已经替换left join为，join因为where它有效地使您的加入inner如果您仍然需要左联接，请将谓词移动到ON子句或在 where 处处理 NULL。

小提琴

于 2019-12-26T20:42:22.410 回答
0

似乎这是解决方案：

select sum(c.impressions_n) / count(1) as average_num_of_impressions from (
select count(i.event) as impressions_n 
  from add4ad i 
  join add4ad c
     on i.user_id = c.user_id and i.unit_id = c.unit_id
 where i.event in ('impression') and c.event in ('create_ad')
 group by i.user_id, i.unit_id              ) c
于 2019-12-26T21:09:04.607 回答
0

问题是检查你必须使用的 NULLS 是 NULL 还是不是 NULL。您在小提琴中的数据也不正确。它对小提琴中的 1234 没有印象。

select i.user_id, i.unit_id, count(i.event) as impressions_n,
count(c.event) as ads_n
from add4ad i
 left join add4ad c
  on i.user_id = c.user_id and i.unit_id = c.unit_id
where i.event in ('impression')
/*and c.event in ('create_ad')*/ and c.spend is not NULL
group by i.user_id, i.unit_id
于 2019-12-26T20:26:22.303 回答