Question

我有一个问题：

SELECT bar, (SELECT name FROM names WHERE value = bar) as name
FROM foobar WHERE foo = 1 and bar = ANY (1,2,3)

我的问题是，当表bar = 3中没有包含foobar（或请求的任何其他值）的行时，不会为该值的bar返回任何行。

我希望我的查询返回[bar, NULL]行，但无法想办法解决这个问题。

这甚至可能吗？

Answer 1

也许这种方法就像你所追求的那样：

测试平台：

create view names as 
select 1 as value, 'Adam' as name union all select 2, 'Beth';

create view foobar as 
select 1 as foo, 1 as bar union all select 1, 2;

原始方法：

select bar, (select name from names where value = bar) as name 
from foobar 
where foo = 1 and bar = any (array[1, 2, 3]);

 bar | name
-----+------
   1 | Adam
   2 | Beth
(2 rows)

替代方法：

with w as (select unnest(array[1, 2, 3]) as bar)
select bar, (select name from names where value = bar) as name
from w left outer join foobar using(bar);

 bar | name
-----+------
   1 | Adam
   2 | Beth
   3 |
(3 rows)

如果您使用的是8.3或之前，则没有内置的unnest功能，但您可以自己动手（非常有效）替换：

create or replace function unnest(anyarray) returns setof anyelement as $$
  select $1[i] from generate_series(array_lower($1,1), array_upper($1,1)) i;
$$ language 'sql' immutable;

Answer 2

SELECT bar, name
FROM foobar
INNER JOIN names ON foobar.bar = names.value
WHERE foo = 1 and bar = ANY (1,2,3)

请尝试该查询。

Answer 3

SELECT  vals.bar, name
FROM    (
        SELECT  *
        FROM    unnest([1, 2, 3]) AS bar
        ) vals
LEFT JOIN
        foobar
ON      foobar.foo = 1
        AND foobar.bar = vals.bar
LEFT JOIN
        names
ON      names.value = vals.bar

在WHERE子句中进行Postgres数组查找

3 个答案: