Question

让我们说person表格中包含name和age列。

我正在编写生成以下SQL的DSL：

select *
  from (select * from person p1 inner join person p2 on p1.name = p2.name) as pj;

现在，我希望能够在外部查询中访问p1和p2，如下所示：

select *
  from (select * from person p1 inner join person p2 on p1.name = p2.name) as pj
  where p1.name = 'xxx'; <-- DOESN'T WORK

像pj.p1.name这样的东西是理想的。如果我不知道person的确切列名，有没有办法实现这一目标？

Answer 1

使用using，然后使用pj.name，或者只使用name

create table person (id serial, name text);
insert into person (name) values ('John'),('Mary');

select *
from (
    select *
    from
        person p1
        inner join
        person p2 using(name)
) r
where name = 'John'
;
 name | id | id 
------+----+----
 John |  1 |  1

USING（a，b，...）形式的一个子句是ON left_table.a = right_table.a和left_table.b = right_table.b ....的简写。另外，USING意味着每个只有一个一对等效列将包含在连接输出中，而不是两者。

如果只需要join的一侧：

select *
from (
    select p1.*
    from
        person p1
        inner join
        person p2 using(name)
) r
where name = 'John'
;
 id | name 
----+------
  1 | John

如果需要join两个边，则使用记录：

select (p2).id, (p1).name -- or (p2).*, (p1).*
from (
    select p1, p2
    from
        person p1
        inner join
        person p2 using(name)
) r
where (p1).name = 'John'
;
 id | name 
----+------
  1 | John

Answer 2

获取额外列为其提供正确的别名

select *
from (select *, 
      p1.name as pjName  --Added Column and use this in where clause 
      from person p1 
      inner join person p2 on p1.name = p2.name) as pj
where pjName = 'xxx';

Answer 3

别名p1和p2不在连接条件的范围内。您只能使用pj别名。您需要做的是在子查询中更明确地说明您选择的列，而不是简单地执行SELECT * ...：

类似的东西：

select *
  from (select p1.name, <add other required columns here> -- explicitly select one of the name columns here
          from person p1 
         inner join person p2 
            on p1.name = p2.name) as pj
  where pj.name = 'xxx' -- use pj alias here.

子查询中的引用别名

3 个答案: