COUNT WHERE 在同一列中满足两个条件

Question

我确定以前有人问过这个问题，但我找不到类似的情况：

给定一个 table 'activity'

user_ID | contact_channel_ID
123456    email
123456    email
555555    SEM
555555    SEM
995959    phone
995959    SEM
696969    email
696969    email

我需要计算拥有 'SEM' 和至少另一个频道 contact_channel_ID

的所有用户

Answer 1

您可以使用聚合和 having:

select user_id
from activity
group by user_id
having sum(case when contact_channel_ID = 'SEM' then 1 else 0 end) > 0 and
       sum(case when contact_channel_ID <> 'SEM' then 1 else 0 end) > 0;

在MySQL中，having子句可以缩写为：

having sum(contact_channel_ID = 'SEM') > 0 and
       sum(contact_channel_ID <> 'SEM') > 0;

Answer 2

您可以使用聚合：

select user_id
from activity
group by user_id
having max(contact_channel_id =  'SEM') = 1 
   and max(contact_channel_id <> 'SEM') = 1

如果要统计这样的用户，再增加一层聚合：

select count(*) as cnt
from (
    select 1
    from activity
    group by user_id
    having max(contact_channel_id =  'SEM') = 1 
       and max(contact_channel_id <> 'SEM') = 1
) t

Answer 3

按用户分组并在HAVING子句中设置条件：

SELECT user_id
FROM activity
GROUP BY user_id
HAVING COUNT(DISTINCT contact_channel_id) > 1
   AND SUM(contact_channel_id = 'SEM') > 0

或使用 EXISTS:

SELECT DISTINCT a.user_id
FROM activity a
WHERE a.contact_channel_id = 'SEM'
AND EXISTS (SELECT 1 FROM activity WHERE user_id = a.user_id AND contact_channel_id <> a.contact_channel_id)

如果你想计算用户数，请将上面的查询更改为：

SELECT COUNT(DISTINCT a.user_id) counter
FROM activity a
WHERE a.contact_channel_id = 'SEM'
AND EXISTS (SELECT 1 FROM activity WHERE user_id = a.user_id AND contact_channel_id <> a.contact_channel_id)

参见demo。
结果：

> | user_id |
> | ------: |
> |  995959 |

和：

> | counter |
> | ------: |
> |       1 |

COUNT WHERE 在同一列中满足两个条件

COUNT WHERE two conditions are met in the same column

mysql

sql

count

having-clause