您可以使用条件聚合来处理这个问题。对于同时使用a和b平台的客户的第一个要求：

SELECT
    Date,
    customer_id AS customerID_active_both_platforms
FROM yourTable
WHERE
    platform IN ('A', 'B')
GROUP BY
    Date,
    customer_id
HAVING
    MIN(platform) <> MAX(platform);

对于第二个要求，既没有平台a也没有平台b的客户：

SELECT
    Date,
    customer_id AS customerID_active_neither_platforms
FROM yourTable
GROUP BY
    Date,
    customer_id
HAVING
    COUNT(CASE WHEN platform IN ('A', 'B') THEN 1 END) = 0;

如果希望每个客户都有此信息，请使用两个聚合级别—首先按客户/日期，然后按客户：

select date,
       sum( is_a * is_b ) as num_ab,
       sum( (1 - is_a) * (1 - is-b) ) as num_not_ab
from (select customer, date,
             max(case when platform = 'A' then 1 else 0 end) as is_a,
             max(case when platform = 'B' then 1 else 0 end) as is_b,
             max(case when platform not in ('A', 'B') then 1 else 0 end) as is_not_ab
      from t
      group by customer, date
     ) t
from t
group by date;

当然，你可以用 where 条款以获得只满足一个或另一个条件的客户。