我在PostgreSQL 8.3中有一个简单的SQL查询,它抓取了一堆注释。我在WHERE子句中为IN构造提供了一个排序的值列表:

SELECT * FROM comments WHERE (comments.id IN (1,3,2,4));

这将以任意顺序返回注释,在my中恰好是id,如1,2,3,4。

我希望结果行像in结构中的列表一样排序:(1,3,2,4)。 如何实现这一目标?


当前回答

你可以很容易地使用(在PostgreSQL 8.2中引入)VALUES(),()。

语法是这样的:

select c.*
from comments c
join (
  values
    (1,1),
    (3,2),
    (2,3),
    (4,4)
) as x (id, ordering) on c.id = x.id
order by x.ordering

其他回答

你可以很容易地使用(在PostgreSQL 8.2中引入)VALUES(),()。

语法是这样的:

select c.*
from comments c
join (
  values
    (1,1),
    (3,2),
    (2,3),
    (4,4)
) as x (id, ordering) on c.id = x.id
order by x.ordering

通过进一步研究,我发现了这个解决方案:

SELECT * FROM "comments" WHERE ("comments"."id" IN (1,3,2,4)) 
ORDER BY CASE "comments"."id"
WHEN 1 THEN 1
WHEN 3 THEN 2
WHEN 2 THEN 3
WHEN 4 THEN 4
END

然而,这似乎相当冗长,并且对于大型数据集可能会有性能问题。 有人能就这些问题发表评论吗?

在Postgres 9.4中,这可以做得更短一些:

select c.*
from comments c
join (
  select *
  from unnest(array[43,47,42]) with ordinality
) as x (id, ordering) on c.id = x.id
order by x.ordering;

或者在没有派生表的情况下更加紧凑:

select c.*
from comments c
  join unnest(array[43,47,42]) with ordinality as x (id, ordering) 
    on c.id = x.id
order by x.ordering

无需手动为每个值分配/维护位置。

在Postgres 9.6中,可以使用array_position():

with x (id_list) as (
  values (array[42,48,43])
)
select c.*
from comments c, x
where id = any (x.id_list)
order by array_position(x.id_list, c.id);

使用CTE时,值列表只需要指定一次。如果这并不重要,也可以写成:

select c.*
from comments c
where id in (42,48,43)
order by array_position(array[42,48,43], c.id);
SELECT * FROM "comments" JOIN (
  SELECT 1 as "id",1 as "order" UNION ALL 
  SELECT 3,2 UNION ALL SELECT 2,3 UNION ALL SELECT 4,4
) j ON "comments"."id" = j."id" ORDER BY j.ORDER

或者如果你喜欢恶而不是善:

SELECT * FROM "comments" WHERE ("comments"."id" IN (1,3,2,4))
ORDER BY POSITION(','+"comments"."id"+',' IN ',1,3,2,4,')
create sequence serial start 1;

select * from comments c
join (select unnest(ARRAY[1,3,2,4]) as id, nextval('serial') as id_sorter) x
on x.id = c.id
order by x.id_sorter;

drop sequence serial;

(编辑)

Unnest在8.3中还没有内置,但你可以自己创建一个(任何的美丽*):

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

该函数可以在任何类型下工作:

select unnest(array['John','Paul','George','Ringo']) as beatle
select unnest(array[1,3,2,4]) as id