在Oracle SQL

Question

脚本1

select supplier.suppl_no, name, count(art_no) TotalArticles 
from article, supplier 
where article.suppl_no = supplier.suppl_no and supplier.suppl_no = 20009
group by supplier.suppl_no, name;

结果：TotalArticles = 51

脚本2

select supplier.suppl_no, name, count(ord_no) TotalOrders 
from orders, supplier 
where orders.suppl_no = supplier.suppl_no and supplier.suppl_no = 20009
group by supplier.suppl_no, name;

结果：TotalOrders = 74

现在，我想将上面两个脚本合并为一个，如下所示。

脚本3

select article.suppl_no, name, count(art_no) "TotalArticles", count(ord_no) "TotalOrders" 
from article, supplier, orders 
where article.suppl_no = supplier.suppl_no and supplier.suppl_no = orders.suppl_no and supplier.suppl_no = 20009
group by article.suppl_no, name

结果：TotalArticles = 3774（预期51与第一个脚本一样）
结果：TotalOrders = 3774（如第二个脚本中预期的74）

所以，在这里，对于TotalArticles和TotalOrders，它给了我51 * 74 = 3774的结果。你能帮我弄清楚我在脚本3中做了什么错吗？非常感谢。

Answer 1

您有一对多关系，因此请加入多重行。使用distinct：

select a.suppl_no, name, count(distinct art_no) "TotalArticles", count(distinct ord_no) "TotalOrders" 
from article a inner join supplier s on (a.suppl_no = s.suppl_no) inner join orders o on (s.suppl_no = o.suppl_no)
where s.suppl_no = 20009
group by a.suppl_no, name;

或者在with声明中生成计数：

with co as (
  select s.suppl_no, name, count(ord_no) TotalOrders 
    from orders o inner join supplier s on (o.suppl_no = s.suppl_no)
    where s.suppl_no = 20009
    group by s.suppl_no, name),
ca as (
  select s.suppl_no, name, count(art_no) TotalArticles 
    from article a inner join supplier s on (a.suppl_no = s.suppl_no)
    where s.suppl_no = 20009
    group by s.suppl_no, name)
select s.suppl_no, ca.TotalArticles, co.TotalOrders 
  from supplier, ca, co 
  where s.suppl_no = 20009;

在两者中你应该使用隐式连接，别名也使查询更具可读性

Answer 2

试试这个。

        select a.suppl_no,name,
        count(a.art_no) "TotalArticles", 
        count(o.ord_no) "TotalOrders"
        from article a 
        inner join on  supplier s on a.suppl_no = s.suppl_no and s.suppl_no = 20009
        inner join on  orders o on o.suppl_no = s.suppl_no  
        group by article.suppl_no, name