加入两个子查询 select 语句不起作用

Question

我正在尝试加入两个 select 语句/查询，以获得结合结果的数据集。 这两个查询分别工作得很好，但是当我尝试使用 join 组合它们时，会发生错误。 错误：第 10 行的“select”处或附近的语法错误，position 1。有人可以帮我，谢谢！

select * from
(
  select dt, count(*) over (order by dt asc) as cumuA from
     (Select date_trunc('week',block_time) AS dt,"from" as sender 
     FROM data where "to" = 'xxxx') 
   as AAA

join 

    Select dt, count(*) over (order by dt asc) as cumuB from ( 
      Select min(date_trunc('week',block_time)) dt, "from" as sender from data where "to" = 'xxxx'
       group by sender)
    as BBB

ON AAA.dt = BBB.dt
)

Answer 1

我不是在看查询实际做了什么，或者是否有更好的方法：只是解决报告的错误。

SELECT * FROM 
(
SELECT * 
 FROM (SELECT dt, count(*) over (order by dt asc) as cumuA 
       FROM (SELECT date_trunc('week',block_time) AS dt,"from" as sender 
             FROM data 
             WHERE "to" = 'xxxx'
            ) as aa
      ) as AAA
join 

SELECT * FROM --< this line makes no sense.  One need not select again after the join. the inline view BBB is sufficient. and the select is just a syntax error.  

(SELECT dt, count(*) over (order by dt asc) as cumuB 
      FROM (SELECT min(date_trunc('week',block_time)) dt, "from" as sender 
            FROM data 
            WHERE "to" = 'xxxx'
            GROUP BY sender
           ) AS bb
      ) as BBB

  ON AAA.dt = BBB.dt
)

我认为你需要它是：

SELECT * FROM 
(
SELECT * 
 FROM (SELECT dt, count(*) over (order by dt asc) as cumuA 
       FROM (SELECT date_trunc('week',block_time) AS dt,"from" as sender 
             FROM data 
             WHERE "to" = 'xxxx'
            ) as aa
      ) as AAA
join 
(SELECT dt, count(*) over (order by dt asc) as cumuB 
      FROM (SELECT min(date_trunc('week',block_time)) dt, "from" as sender 
            FROM data 
            WHERE "to" = 'xxxx'
            GROUP BY sender
           ) AS bb
      ) as BBB

  ON AAA.dt = BBB.dt
)