Question

如何从有子句中的子选择中引用网络列？

select distinct c.id, c.name,
(
select count(cm.id) cnt
from
company_mapping cm
where
cm.company_id_source = c.id
or
cm.company_id_target = c.id
) network
from company c
where
c.name like 'foobar%'
group by c.id, c.name
having network > 1

ORA-00904：“NETWORK”：标识符无效。如果我遗漏了最后一行，它按预期工作，但我只对网络＆gt;行感兴趣1。

Answer 1

您无法访问select，group by或having中where中定义的字段。

sql运算符的顺序如下：

1.FROM clause
2.WHERE clause
3.GROUP BY clause
4.HAVING clause
5.SELECT clause
6.ORDER BY clause

这就是为什么你可以在network中使用order by但在select之前的运算符中不能使用{。}}。

Answer 2

你想做这样的事吗？

select c.id
      ,c.name
      ,count(*) 
      ,count(s.company_id_source) as num_sources
      ,count(t.company_id_target) as num_targets
  from company              c
  left join company_mapping s on(s.company_id_source = c.id)
  left join company_mapping t on(t.company_id_target = c.id)
 where c.name like 'foobar%'  
 group 
    by c.id
      ,c.name
having count(s.company_id_source) > 1
    or count(t.company_id_target) > 1;

修改：下面的新查询以回应评论。该查询现在返回：所有匹配“Foobar”的公司，无论他们是否在表company_mapping中都有关联的行，以及：

num_sources：公司所在的company_mapping中的nr行来源。

num_targets：company_mapping中行的nr 公司是目标。

num_mappings：company_mapping（源或目标）中的“连接”nr

select c.id ,c.name ,count(s.company_id_source) + count(t.company_id_target) as num_mappings ,count(s.company_id_source) as num_sources ,count(t.company_id_target) as num_targets from company c left join company_mapping s on(s.company_id_source = c.id) left join company_mapping t on(t.company_id_target = c.id) where c.name like 'foobar%' group by c.id ,c.name;

Answer 3

Ronnis在正确的道路上，但查询实际上应该更简单。请尽量避免选择内部选择，因为它在99％的时间都是性能杀手。

select c.id
,      c.name
,      count(*) network
from   company c
join   company_mapping cm on c.id in (cm.company_id_source, cm.company_id_target)
where  c.name like 'foobar%'
group by c.id, c.name
having count(*) > 1

Answer 4

~~首先，你不能在同一个查询中拥有distinct和group by。~~这只是多余的，你是对的，我不知道为什么oracle没有'抛出异常。

其次，别名与查询在同一级别是未知的。你应该将它包含在外部查询中。

select id, name, network
from (
    select c.id, c.name,
      (
      select count(cm.id) cnt
      from
      company_mapping cm
      where
      cm.company_id_source = c.id
      or
      cm.company_id_target = c.id
      ) network
    from company c
    where
    c.name like 'foobar%'
    group by c.id, c.name
)
WHERE network > 1;

Oracle：在having子句中命名子查询

4 个答案: