BigQuery 避免多个子查询

Question

我们正在开发一个应用程序，它将请求存储在一个表中，并将响应存储在另一个表中（当然）。每个请求可以有多个响应，并且我们将请求 ID 存储在两个表中。

最初，我认为我们可以使用请求中的左连接 -> 响应来计算每个匹配条件的总数：

SELECT source, COUNT(*) as requests, COUNT(responses.request_id) as responses
FROM DATASET.requests
LEFT JOIN DATASET.responses ON requests.id = responses.request_id
WHERE source = "source1"
GROUP BY source

有 70 个请求符合 WHERE 条件，30 个响应符合这个条件。预期输出为：“source1, 70, 30”。 从那以后，我了解了更多关于 JOIN 行为的信息，而我们得到了“source1, 259, 207”。两边有重复的ID。

我能够得到我想要的结果的唯一方法是创建一个巨大的查询，以及多个在给定条件过滤的 ID 集中匹配的完整子查询。然后使用过滤后的 ID 集来真正提取我们的字段、统计信息等。

SELECT * FROM
  (SELECT COUNT(*) as responses FROM DATASET.responses
  WHERE id IN (SELECT id FROM DATASET.requests WHERE source = 
  "source1"))
  ,
 (SELECT source, COUNT(*) as requests
  FROM  PUBDATA.requests
  WHERE id IN (SELECT id FROM DATASET.requests WHERE source = "source1")
  GROUP BY source)

这看起来很糟糕。我曾尝试使用 CTE 来收集我们想要的 ID 列表，并使用 WHERE id/request_id IN (cte.id) 但这显然是不可能的，除非我们在 cte 上加入，这再次产生错误和成倍的结果。

由于我们想在查询中添加额外的统计信息，这将需要更多的 WHERE 子句，我担心这个怪物会继续增长并且难以实现。

如果有更好的方法，请告诉我。谢谢！

编辑 - 请求的示例架构 请求

id (String), source (String), partner_ids (Integer array), user_agent (String), timestamp (Timestamp), ...

回复

request_id (String, from requests.id), partner_id (Integer), is_billed (boolean), price_charged (float, null if is_billed = false), response_categories (String array, not from requests), ...

挑战在于我们必须主要查询 Requests 表以获取与我们的条件匹配的 ID 值列表，然后查询每个表的统计信息（例如 counts、count where is_billed 等）以获取一份合并报告。我们可能还需要从每个表的条件中提取 ID 池（例如 where requests.source = 'source1' 和response.response_categories IN 'action'）

Answer 1

也许我误会了什么，你为什么不只计算每个并加入 id 呢？

WITH
    sources
    AS
        (  SELECT COUNT (*) source_cnt, id
             FROM dataset.request
         GROUP BY id),
    responses
    AS
        (  SELECT COUNT (*) AS response_cnt, id
             FROM dataset.responses
         GROUP BY id)
SELECT source_cnt, response_cnt, sources.id
  FROM sources INNER JOIN responses ON sources.id = responses.id;

如果您想保留所有记录，可以将其修改为完全外部联接：

WITH
    sources
    AS
        (  SELECT COUNT (*) source_cnt, id
             FROM dataset.request
         GROUP BY id),
    responses
    AS
        (  SELECT COUNT (*) AS response_cnt, id
             FROM dataset.responses
         GROUP BY id)
SELECT COALESCE (sources.id, responses.id) AS id, source_cnt, response_cnt
  FROM sources FULL OUTER JOIN responses ON sources.id = responses.id

Answer 2

我认为你可以用union all 和group by 做你想做的事：

select source, sum(requests) as requests, sum(responses) as responses
from ((select source, count(*) as requests, 0 as response
       from dataset.requests
       group by source
      ) union all
      (select source, 0 as requests, count(*) as responses
       from dataset.responses
       group by source
      )
     ) rr
group by source;

这会对所有来源进行计算。

编辑：

对于修改后的版本，只需多加一个join：

select source, sum(requests) as requests, sum(responses) as responses
from ((select source, count(*) as requests, 0 as response
       from dataset.requests rq
       group by rq.source
      ) union all
      (select rq.source, 0 as requests, count(*) as responses
       from dataset.responses r join
            (select distinct rq.id
             from dataset.requests rq
            ) rq
            on r.id = rq.id
       group by rq.source
      )
     ) rr
group by source;

如果每个请求最多有一个响应，您可以将其缩短为：

select rq.source, count(*) as requests, count(r.id) as responses
from dataset.requests rq left join
     dataset.responses r
     on r.id = rq.id
group by rq.source

Answer 3

老实说，我对您最终希望看到的内容感到有些困惑，而且我也不完全理解如果一个请求可以有多个响应，那么您如何有 70 个请求而只有 30 个响应。您的意思是某些请求可以有 0 个响应吗？还是您在计算不同的响应？

如果您希望计算请求总数以及与这些特定请求相关的响应总数，我相信对您的代码进行的这种轻微修改应该可以工作：

SELECT source, COUNT(DISTINCT id) as requests, COUNT(responses.request_id) as responses
FROM `dataset.requests` as requests
LEFT JOIN `dataset.responses` as responses ON requests.id = responses.request_id
WHERE source = "source1"
GROUP BY source