聚合函数在连接更多表时返回错误的值

我想显示所有的客户和他们的地址,以及他们的订单数量和总金额。我的查询如下所示:

select *, sum(o.tota), count(o.total) 
from customer c 
natural join orders o
group by c.custId;

select *, sum(o.tota), count(o.total) 
from customer c 
natural join orders o
natural join cust_addresses a
group by c.custId;

那就不管用了。聚合函数返回错误的值,因为每个客户可能有多个地址,这是正确的,我还想显示它们的所有地址。

select *, (select total from orders o where o.custid=c.custid), ..
from customer c 
natural join orders o
natural join cust_addresses a
group by c.custId;

但这很慢。

我现在尝试了以下操作,但它告诉我字段c.custid未知:

select *
from
     customer c,               
     left join (select sum(o.tota), count(o.total) from orders o where o.custid=c.custid) as o
where ...
group by c.custId;