Question

我已将所有内容别名，但似乎无法在没有

的情况下运行此查询

错误列引用\＆＃34; id \＆＃34;是不明确的＆＃34;

如果我删除其中一个联接似乎有效，但我只是对为什么它不能与两个人合作感到困惑？

create function influence.person_info(user_id integer)
  returns setof influence.person_object as
$$
    declare
    obj influence.person_object;
    begin
        select t1.email as a_email, t2.organisation_url as a_org, t3.first_name as a_first, t3.last_name as a_last into obj
        from influence_private.person_account as t1
        inner join influence_private.organisation_account as t2 on (t1.organisation_id = t2.id)
        inner join influence.person as t3 on (t1.person_id = t3.id)
        where id = $1;
        return next obj;
    end;
$$ LANGUAGE plpgsql stable;

任何指针？

Answer 1

如果您查看查询：

select t1.email as a_email, ...
from influence_private.person_account as t1
join influence_private.organisation_account as t2 
  on (t1.organisation_id = t2.id)   -- here
join influence.person as t3 
  on (t1.person_id = t3.id)         -- here

您的表格（influence_private和influence）都有名为id的列。 Postgresql不知道你要使用哪一个，所以你必须使用像t2.id这样的全名。

Answer 2

在 where 子句中，您需要明确说明您所指的 id 。

Answer 3

您的查询问题是t2和t3两个表都有一个名为id的列。在Where条款中id是不明确的，因为它不知道您引用的id（t2或t3）指定了它，它会正常工作

示例修复：

create function influence.person_info(user_id integer)
  returns setof influence.person_object as
$$
    declare
    obj influence.person_object;
    begin
        select t1.email as a_email, t2.organisation_url as a_org, t3.first_name as a_first, t3.last_name as a_last into obj
        from influence_private.person_account as t1
        inner join influence_private.organisation_account as t2 on (t1.organisation_id = t2.id)
        inner join influence.person as t3 on (t1.person_id = t3.id)
        where t1.id = $1;
        return next obj;
    end;
$$ LANGUAGE plpgsql stable;

PostgreSQL - 错误列参考＆＃34; id＆＃34;很暧昧

3 个答案: