Question

我有表watchers：

------------------------------------------
time_type | time_count | timestamp_created
------------------------------------------
  'hour'  |   1        |   2016-12-08 15:56:26.169614
  'hour'  |   13       |   ... 
  'day'   |   5        |   ...

我尝试根据time_type和time_count值获取整数：

CREATE OR REPLACE FUNCTION af_calculate_range(tt text,tc  integer) RETURNS integer AS $$ 
            SELECT tt, CASE tt WHEN 'day' THEN tc * 60 * 60 
                               WHEN 'hour' THEN tc * 60
                       END
        FROM watchers;

          $$
    LANGUAGE SQL;

SELECT af_calculate_range(time_type, time_count) FROM watchers

但是我收到了错误：

ERROR:  return type mismatch in function declared to return integer  
DETAIL:  Final statement must return exactly one column. 
CONTEXT:  SQL function "af_calculate_range"
********** Error **********

用法：

SELECT * FROM watchers
WHERE  EXTRACT(EPOCH FROM now()) > (EXTRACT(EPOCH FROM timestamp_created) +
 af_calculate_range(time_type, time_count) )

如果time_type = 'hour'且time_count = 1，则输出应为3600秒。

我的例子出了什么问题：

我使用了https://www.postgresql.org/docs/7.4/static/functions-conditional.html

和https://www.postgresql.org/docs/9.1/static/sql-createfunction.html

Answer 1

一个函数只能返回一个值，所以你可能打算这样做：

CREATE OR REPLACE FUNCTION af_calculate_range(tt text, tc integer) RETURNS integer AS $$ 
        SELECT CASE WHEN tt = 'day' THEN tc * 60 * 60    -- return a (single) scalar 
                    WHEN tt = 'hour' THEN tc * 60
               END
        FROM watchers;
      $$
LANGUAGE SQL;

但正如@Mureinik指出的那样，你甚至不需要做SELECT;只需直接使用CASE表达式。

Answer 2

您不需要在那里查询，因为您将所有值传递给函数 - 只需返回case的值：

CREATE OR REPLACE FUNCTION af_calculate_range(tt TEXT, tc INTEGER) 
RETURNS INTEGER IMMUTABLE AS $$
BEGIN
    RETURN CASE tt WHEN 'day' THEN tc * 60 * 60
                   WHEN 'hour' THEN tc * 60
           END;
END;
$$
LANGUAGE PLPGSQL;

Answer 3

你太复杂了。存储interval代替

user_id

现在查询将是：

select now(), now() + interval '2 days';
              now              |           ?column?            
-------------------------------+-------------------------------
 2016-12-19 05:23:51.856137-02 | 2016-12-21 05:23:51.856137-02

create table watchers (
    the_interval interval,
    timestamp_created timestamp
);

insert into watchers (the_interval, timestamp_created) values
('1 hour', '2016-12-08 15:56:26.169614');

Postgres定制功能与CASE

3 个答案: