mysql学习打卡day25

今日成果：

select * 
from clients
where client_id in (
select client_id
from invoices
group by client_id
having count(*) >=2
);
-- 另一种写法
select * 
from clients
where client_id = any (
select client_id
from invoices
group by client_id
having count(*) >=2
);
-- 查询大于2张发票的用户

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