SQL

Question

我正在尝试根据列中的值将结果集中的列拆分为 2 列。因此，一个用户可以订阅多个项目，并且该用户可以有 2 个可以接收此订阅的电子邮件地址。结果集给出了订阅列表及其对应的订阅电子邮件 ID 条目。

数据库详细信息

Table 1 - user_subscriptions

user_id 
email_id - 1 for email id 1 and 2 for email id 2
subscription_id

Table 2 - subscriptions 

subscription_id
subscription_name

现在我需要用户的所有订阅，无论是否通过任一电子邮件 ID 订阅。所以我得到了这样的结果集

+----------------------+----------+
| subscription_name    | email_id |
+----------------------+----------+
| item1                | 1        |
| item1                | 2        |
| item2                | null     |
| item3                | 1        |
| item4                | null     |
| item5                | 2        |
+----------------------+----------+

所以我想将上面的结果集拆分成如下的内容

+-------------------+---------+---------+
| subscription_name | email_1 | email_2 |

+-------------------+---------+---------+
| item1             | 1 or Y  | 1 or Y  |
| item2             | 0 or N  | 0       |
| item3             | 1       | 0       |
| item4             | 0       | 0       |
| item5             | 0       | 1       |
+-------------------+---------+---------+

希望这个问题是有道理的。如有任何帮助，我们将不胜感激！

已更新------------

示例数据：

subscriptions - +-----------------+-------------------+ | subscription_id | subscription_name | +-----------------+-------------------+ | 1 | item1 | | 2 | item2 | | 3 | item3 | | 4 | item4 | | 5 | item5 | +-----------------+-------------------+

user_subscriptions

+---------+----------+-----------------+ | user_id | email_id | subscription_id | +---------+----------+-----------------+ | 101 | 1 | 1 | | 101 | 2 | 1 | | 101 | 1 | 3 | | 101 | 2 | 5 | | 102 | 1 | 1 | | 102 | 2 | 1 | +---------+----------+-----------------+

预期结果：

对于user_id = 101

+-----------------+-------------------+--------+--------+ | subscription_id | subscription_name | mail_1 | mail_2 | +-----------------+-------------------+--------+--------+ | 1 | item1 | Y | Y | | 2 | item2 | N | N | | 3 | item3 | Y | N | | 4 | item4 | N | N | | 5 | item5 | N | Y | +-----------------+-------------------+--------+--------+

Answer 1

您需要一个条件聚合：

select us.subscription_name,
     -- there's at least one email
   CASE WHEN MIN(us.email_id) IS NOT NULL THEN 'Y' ELSE 'N' END as email_1,
     -- there's more than one email
   CASE WHEN MIN(us.email_id) <> MAX(us.email_id) THEN 'Y' ELSE 'N' END as email_2
from subscriptions as s
left join user_subscriptions as us
on s.subscription_id = us.subscription_id
where us.user_id = ...  
group by us.subscription_name

Answer 2

SELECT 
     S.subscription_id, 
     S.subscription_name,
     CASE 
         WHEN US1.mail_ID IS NULL THEN 'N'
         ELSE 'Y'
     END mail_1,
     CASE 
         WHEN US2.mail_ID IS NULL THEN 'N'
         ELSE 'Y'
     END mail_2
FROM subscriptions S
LEFT JOIN user_subscriptions US1
  ON S.subscription_id = US1.subscription_id
 AND US1.mail_id = 1
LEFT JOIN user_subscriptions US2
  ON S.subscription_id = US2.subscription_id
 AND US2.mail_id = 2
WHERE us1.user_id = 5 -- or use a variable @user_ID
  OR  us2.user_id = 5

Answer 3

我以前没有在 sybase 工作过，但我很确定以下 SQL 会很容易翻译（甚至运行直接翻译）：

SELECT
    s.subscription_name,
    COUNT(email_1.subscription_id) AS email_1,
    COUNT(email_2.subscription_id) AS email_2
FROM subscriptions AS s
LEFT JOIN user_subscriptions AS email_1 ON (
    s.subscription_id = email_1.subscription_id AND
    email_1.email_id = 1
)
LEFT JOIN user_subscriptions AS email_2 ON (
    s.subscription_id = email_2.subscription_id AND
    email_2.email_id = 2
)
;

你也可以在 SELECT 中说 IF(email_1.subscription_id IS NOT NULL, 'Y', 'N') 等到 return 直接 yes/no 而不是计数等

它的工作原理是 LEFT JOIN 语句列表将匹配任何 "user subscription" 记录 email_id=1 和 email_id=2 等

我缺乏 sybase 知识免责声明： ANSI SQL 无法执行 PIVOT - 如果 sybase 可以，您可以更优雅地执行此操作我确定。还有另一个问题+答案暗示sybase可以做这样的事情；值得你去看看：

希望对您有所帮助！

SQL - 根据值拆分列

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