Question

我有两个表，paid_users和paid_users_no还有更多。从本质上讲，付费用户可以拥有多个帐户。

他们的模式如下：

paid_users:
payor_id | user_email | payor_email | payment_start_date
---------------------------------------------------------
         |            |             |


paid_users_no_more:
user_id | payment_stop_date
---------------------------
        |

我写了一个查询来获取用于支付哪些帐户以及在给定月份中哪些帐户不再支付（流失的帐户）：

SELECT payor_id, count(*) as "churned accounts" FROM paid_users_no_more 
INNER JOIN paid_users 
ON paid_users_no_more.user_id=paid_users.user_id 
WHERE paid_users.payment_start_date NOT BETWEEN '2015-08-01 00:00:00'::timestamp AND '2015-08-30 23:59:59'::timestamp 
AND paid_users_no_more.payment_stop_date BETWEEN '2015-08-01 00:00:00'::timestamp AND '2015-08-30 23:59:59'::timestamp 
GROUP BY paid_users.payor_id;

这为我提供了8月份每个payor_id的流失帐户数量 - 我如何获取付款人拥有的帐户总数？即，以下查询为我提供了每个payor_id的帐户数量：

SELECT paid_users.payor_email,count(*) AS "total accounts" 
FROM paid_users 
WHERE paid_users.payment_start_date NOT BETWEEN '2015-08-01 00:00:00'::timestamp AND '2015-08-30 23:59:59'::timestamp 
GROUP BY paid_users.payor_email;

我希望能够以某种方式加入两个结果表来查看＆＃34;搅拌帐户＆＃34;和＆＃34;总帐户＆＃34; （因为我想计算流失的收入，而且会有更多的帐户，它们的成本会越便宜） - 有没有办法加入这些表？

谢谢！

Answer 1

如果您使两个查询Union Compatible，那么您可以将结果和聚合与其他查询结合起来：

    SELECT payer, 
           sum(churned_accounts) AS "churned_count",
           sum(total_accounts) AS "total_count" 
    FROM (
      SELECT CAST(payor_id AS CHAR(50)) AS "payer", 
             count(*) as "churned accounts", 
             0 AS "total accounts" 
      FROM paid_users_no_more 
      INNER JOIN paid_users 
        ON paid_users_no_more.user_id=paid_users.user_id 
      WHERE paid_users.payment_start_date NOT BETWEEN '2015-08-01 00:00:00'::timestamp AND '2015-08-30 23:59:59'::timestamp 
      AND paid_users_no_more.payment_stop_date BETWEEN '2015-08-01 00:00:00'::timestamp AND '2015-08-30 23:59:59'::timestamp 
      GROUP BY paid_users.payor_id

      UNION

      SELECT CAST(paid_users.payor_email AS CHAR(50)) AS "payer", 
             0 AS "churned accounts", 
             count(*) AS "total accounts" 
      FROM paid_users 
      WHERE paid_users.payment_start_date NOT BETWEEN '2015-08-01 00:00:00'::timestamp AND '2015-08-30 23:59:59'::timestamp 
      GROUP BY paid_users.payor_email
    ) as All_Accounts

拥有0 AS "total_accounts"和0 AS "churned_accounts"表示两个查询具有相同的字段，这使UNION成为可能。

如何获取INNER联接的计数和所有项目的计数？

1 个答案: