Mysql 引用嵌套查询的派生表

发布于 2024-10-22 19:10:29 字数 733 浏览 8 评论 0原文

我昨天发布了与此类似的内容,但现在我想要与我的查询稍有不同的内容 -

我正在尝试查询数据库以检索一段时间内访问过某个网站的一次性用户的数量。数据看起来像这样:

Day | UserID  
  1 | A  
  1 | B  
  2 | B  
  3 | A  
  4 | B  
  4 | C  
  5 | D  

我希望查询结果看起来像这样

Time Span      | COUNT(DISTINCT UserID)  
Day 1 to Day 1 | 2  
Day 1 to Day 2 | 1  
Day 1 to Day 3 | 0  
Day 1 to Day 4 | 1  
Day 1 to Day 5 | 2  

结果是 2,1,0,1,2 因为,在这些天结束时,有 X 个用户拥有访问过一次。例如,对于第 5 天,在第 5 天结束时,用户 c 和 d 仅访问过一次。

我想我正在寻找与此类似的查询:

select d.day, (select count(distinct userid) from visits where day<=d.day)
from (select distinct day from visits) d

上面的查询与我的查询之间的差异寻找的是我希望这个新查询只考虑每个时间跨度的一次性用户,而不是重复用户。

谢谢

I posted something similar to this yesterday, but now I'd like something a little different from my query-

I'm trying to query a database to retrieve the number of one-time users who have visited a website over time. The data looks something like this:

Day | UserID  
  1 | A  
  1 | B  
  2 | B  
  3 | A  
  4 | B  
  4 | C  
  5 | D  

I'd like the query result to look this this

Time Span      | COUNT(DISTINCT UserID)  
Day 1 to Day 1 | 2  
Day 1 to Day 2 | 1  
Day 1 to Day 3 | 0  
Day 1 to Day 4 | 1  
Day 1 to Day 5 | 2  

The result is 2,1,0,1,2 because, at the end of those days, there are X number of users who have visited a single time. e.g. for day 5, at the end of day 5, users c and d have visited only once each.

I think I'm looking for a query similar to this:

select d.day, (select count(distinct userid) from visits where day<=d.day)
from (select distinct day from visits) d

The difference between the query above and what I'm looking for is that I'd like this new query to consider only one-time users for each time span, and not repeat users.

Thanks

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甜点 2024-10-29 19:10:29

该子查询应该适用于明确的要求。

select d.day, count(distinct case when b.userid is null then a.userid end)
from (select day from visits group by day) d
inner join
(
select a.day, a.userid, count(*) c
from visits a
join visits b on a.userid=b.userid and b.day <= a.day
group by a.day, a.userid
having count(*) = 1
) a on a.day <= d.day
left join
(
select a.day, a.userid, count(*) c
from visits a
join visits b on a.userid=b.userid and b.day <= a.day
group by a.day, a.userid
having count(*) > 1
) b on a.userid = b.userid and b.day <= d.day
group by d.day

原文

您一定是从 SQL Server 中汲取了这个想法 - 它是唯一允许您引用两次删除(嵌套)查询的 RDBMS (IIRC)。请指出您想要什么,我们可以重写查询。

对于显示的确切查询,您不需要 2 级子查询

SELECT  
    C.col_c1 AS Data,
    (
        SELECT count(col_b1)
        FROM tbl
        WHERE col_b2 <= C.col_c1
    ) A
FROM (
    SELECT col_c1 # subquery to get distinct c1
    FROM tbl
    GROUP BY col_c1) C;

This subquery should work for the clarified requirements.

select d.day, count(distinct case when b.userid is null then a.userid end)
from (select day from visits group by day) d
inner join
(
select a.day, a.userid, count(*) c
from visits a
join visits b on a.userid=b.userid and b.day <= a.day
group by a.day, a.userid
having count(*) = 1
) a on a.day <= d.day
left join
(
select a.day, a.userid, count(*) c
from visits a
join visits b on a.userid=b.userid and b.day <= a.day
group by a.day, a.userid
having count(*) > 1
) b on a.userid = b.userid and b.day <= d.day
group by d.day

Original

You must have taken the idea from SQL Server - it is the only RDBMS (IIRC) that will allow you to reference a twice removed (nesting) query. Please indicate what you want and we can rewrite the query.

For the exact query shown, you don't need 2 levels of subquery

SELECT  
    C.col_c1 AS Data,
    (
        SELECT count(col_b1)
        FROM tbl
        WHERE col_b2 <= C.col_c1
    ) A
FROM (
    SELECT col_c1 # subquery to get distinct c1
    FROM tbl
    GROUP BY col_c1) C;
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