Postgresql 需要不同于postgres函数的返回格式

我有一张桌子

CREATE TABLE reward_transactions
(
  uid bigint NOT NULL,
  reward_type text NOT NULL,
  count bigint,
  last_update timestamp with time zone DEFAULT now()
)

和一个函数

CREATE FUNCTION store_reward_transaction(bigint, text) returns record
    LANGUAGE plpgsql
    AS $_$
declare
_record record;
begin
update reward_transactions set count = count + 1, last_update = now() where uid = $1 and reward_type = $2::text returning count, last_update::timestamp with time zone into _record;
if found then
return _record;
end if;
begin
insert into reward_transactions (uid, count, reward_type) values ($1, 1, $2::text) returning count, last_update::timestamp with time zone into _record;
return _record;
exception when unique_violation then
update reward_transactions set count = count + 1, last_update = now() where uid = $1 and reward_type = $2::text returning count, last_update::timestamp with time zone into _record;
return _record;
end;
end
$_$;

问题是，我如何才能让它不返回

行：[{store_renward_transaction:'（12，“2014-07-18 17:29:39.780207-05”）}

，而是返回类似于从原始表中选择的内容（即列名完整），我已尝试将其作为tablename%ROWTYPE返回，使用自定义类型和其他一些东西，我似乎无法获得所需的输出。使用9.3.3尝试以下操作：

SELECT * FROM store_reward_transaction(4,'blah') f(count bigint, last_update timestamp with time zone);

---

感谢您让我走上了正确的轨道，相应地更新您的答案

select*from store_reward_transaction（4，'blah'）作为f（count bigint，last_update timestamp with time zone）我会标记它，您的原始答案会导致
错误：返回“record”的函数需要列定义列表