方法一

• 连接两个表格，筛选出只回答了教育和职场类的用户
• group by issue_type,author_id使得就算一个用户回答了多个教育或者职场问题，也只算作一个，因为这里是统计用户个数，接着order by author_id，将相同用户放在一起
• 利用窗口函数lead形成新的一列leadd，partition by author_id
• 上面三步形成新的表格t1，再统计leadd个数就好啦

select
count(leadd) as num
from
(select 
	issue_type,
    author_id,
    lead(issue_type,1) over (partition by author_id) as leadd
from issue_tb it
join answer_tb atb
on it.issue_id=atb.issue_id
where issue_type='Education' or issue_type='Career'
group by issue_type,author_id
order by author_id) t1

方法二

• 利用子查询

select 
	count(distinct author_id) as num
from issue_tb it
join answer_tb atb
on it.issue_id=atb.issue_id
where issue_type='Education' 
      and author_id in( select author_id
						from issue_tb it
						join answer_tb atb
						on it.issue_id=atb.issue_id
                        where issue_type='Career' )