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);

这里有另一个解决方案，它使用了一个常量表(http://www.postgresql.org/docs/8.3/interactive/sql-values.html):)

SELECT * FROM comments AS c,
(VALUES (1,1),(3,2),(2,3),(4,4) ) AS t (ord_id,ord)
WHERE (c.id IN (1,3,2,4)) AND (c.id = t.ord_id)
ORDER BY ord

但我不确定这是表现性的。

我现在有一堆答案了。我可以得到一些投票和评论，这样我就知道哪一个是赢家!

谢谢大家:-)

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

只是因为它很难找到，它必须被传播:在mySQL中，这可以做得更简单，但我不知道它是否适用于其他SQL。

SELECT * FROM `comments`
WHERE `comments`.`id` IN ('12','5','3','17')
ORDER BY FIELD(`comments`.`id`,'12','5','3','17')

select * from comments where comments.id in 
(select unnest(ids) from bbs where id=19795) 
order by array_position((select ids from bbs where id=19795),comments.id)

这里，[bbs]是主表，它有一个名为ids的字段， id是存储注释的数组。id。

在postgresql 9.6中通过

让我们对已经说过的内容有一个直观的印象。例如，你有一个带有一些任务的表:

SELECT a.id,a.status,a.description FROM minicloud_tasks as a ORDER BY random();

 id |   status   |   description    
----+------------+------------------
  4 | processing | work on postgres
  6 | deleted    | need some rest
  3 | pending    | garden party
  5 | completed  | work on html

并且您希望按状态对任务列表进行排序。 status是一个字符串值列表:

(processing, pending,  completed, deleted)

诀窍是给每个状态值一个整数，并以数字顺序排列列表:

SELECT a.id,a.status,a.description FROM minicloud_tasks AS a
  JOIN (
    VALUES ('processing', 1), ('pending', 2), ('completed', 3), ('deleted', 4)
  ) AS b (status, id) ON (a.status = b.status)
  ORDER BY b.id ASC;

这就导致:

 id |   status   |   description    
----+------------+------------------
  4 | processing | work on postgres
  3 | pending    | garden party
  5 | completed  | work on html
  6 | deleted    | need some rest

信贷@user80168