解题思路:

1.找出uid不同的但是买过pro_id相同的商品的用户 [自连接]

 selecta.uid,b.pro_id fromtb_order ajoin tb_order bon a.pro_id = b.pro_id where a.uid <> b.uid 	 -- 找出uid不同的但是pro_id相同的商品

2.对uid和pro_id进行分组

selecta.uid,a.pro_id 
fromtb_order a
join tb_order b
on a.pro_id = b.pro_id where a.uid <> b.uid 	 -- 找出uid不同的但是买过pro_id相同的商品的用户
group by uid,pro_id 

3.在对uid进行分组得到买过两件及以上的用户

select 
uid  
from(select uid  from(selecta.uid,a.pro_id fromtb_order ajoin tb_order bon a.pro_id = b.pro_id where a.uid <> b.uid 	 -- 找出uid不同的但是买过pro_id相同的商品的用户group by uid,pro_id ) tb0
group by uid   -- 得到买商品相同数为2件及以上的用户   
having count(1) >= 2

 4.再进行自连接得到符合条件的人和商品

selecttb1.uid,tb1.pro_id
fromtb_order tb1
join(select uid  from(select uid  from(selecta.uid,a.pro_id fromtb_order ajoin tb_order bon a.pro_id = b.pro_id where a.uid <> b.uid 	 -- 找出uid不同的但是买过pro_id相同的商品的用户group by uid,pro_id ) tb0group by uid   -- 得到买商品相同数为2件及以上的用户   having count(1) >= 2) tb2
on tb1.uid = tb2.uid

 5.在对pro_id进行分区获取商品数是1的商品[使用自查询即可]

最终sql:

selecttb3.uid,tb3.pro_id
from(selecttb1.uid,tb1.pro_idfromtb_order tb1join(select uid  from(select uid  from(selecta.uid,a.pro_id fromtb_order ajoin tb_order bon a.pro_id = b.pro_id where a.uid <> b.uid 	 -- 找出uid不同的但是买过pro_id相同的商品的用户group by uid,pro_id ) tb0group by uid   -- 得到买商品相同数为2件及以上的用户   having count(1) >= 2) tb2on tb1.uid = tb2.uid) tb3 
where pro_id in (select pro_id from  tb_order group by pro_id having count(1) = 1)