10

我正在运行一个查询,该查询提供一组不重叠的 first_party_id - 与一个第三方关联但与另一个无关的 ID。但是,此查询不在 Athena 中运行,并给出错误:Correlated queries not yet supported.

正在查看 prestodb 文档, https ://prestodb.io/docs/current/sql/select.html(Athena 是 prestodb 的底层),以寻找嵌套查询的替代方案。给出的 with statement例子似乎不能很好地翻译这个not in子句。想知道嵌套查询的替代方法是什么——下面的查询。

SELECT 
         COUNT(DISTINCT i.third_party_id) AS uniques
FROM 
         db.ids i
WHERE
         i.third_party_type = 'cookie_1'
         AND i.first_party_id NOT IN (
           SELECT
             i.first_party_id
           WHERE 
             i.third_party_id = 'cookie_2'
         )
4

2 回答 2

15

可能有更好的方法来做到这一点 - 我也很想看到它!我能想到的一种方法是使用外连接。(我不太确定您的数据是如何构成的,所以请原谅这个人为的例子,但我希望它可以翻译得很好。)这个怎么样?

with 
  a as (select * 
       from (values 
       (1,'cookie_n',10,'cookie_2'), 
       (2,'cookie_n',11,'cookie_1'),
       (3,'cookie_m',12,'cookie_1'),
       (4,'cookie_m',12,'cookie_1'),
       (5,'cookie_q',13,'cookie_1'),
       (6,'cookie_n',13,'cookie_1'),
       (7,'cookie_m',14,'cookie_3')
       ) as db_ids(first_party_id, first_party_type, third_party_id, third_party_type)
      ),
  b as (select first_party_type 
        from a where third_party_type = 'cookie_2'),
  c as (select a.third_party_id, b.first_party_type as exclude_first_party_type 
        from a left join b on a.first_party_type = b.first_party_type 
        where a.third_party_type = 'cookie_1')
select count(distinct third_party_id) from c 
       where exclude_first_party_type is null;

希望这可以帮助!

于 2017-02-02T15:49:04.100 回答
1

您可以使用外部联接:

SELECT 
         COUNT(DISTINCT i.third_party_id) AS uniques
FROM 
         db.ids a
LEFT JOIN
         db.ids b
ON       a.first_party_id = b.first_party_id
     AND b.third_party_id = 'cookie_2'
WHERE
         a.third_party_type = 'cookie_1'
     AND b.third_party_id is null -- this line means we select only rows where there is no match

NOT IN在使用可能返回值的子查询时,您还应该小心,NULL因为条件将始终为真。您的查询与 进行比较a.first_party_idNULL这将始终为假,因此NOT IN将导致条件始终为真。讨厌的小问题。

避免这种情况的一种方法是避免使用NOT IN或向子查询添加条件,即AND third_party_id IS NOT NULL.

有关详细说明,请参见此处

于 2019-04-01T13:21:29.827 回答