sql - PostgreSQL 函数返回表

Question

我想在 PostgreSQL 上设置一个返回表的函数。这是函数的源代码:

   CREATE OR REPLACE FUNCTION feiertag(inDate Date) 
     RETURNS TABLE (eingabeDatum DATE, f_heute INT, f_1 INT, f_2 INT, f_3 INT, f_5 INT) 
    AS $$
            DECLARE
              f_heute integer := 0;
              f_1 integer := 0;
              f_2 integer := 0;
              f_3 integer := 0;
              f_5 integer := 0;
            BEGIN
                    SELECT 1 INTO f_heute FROM feiertage where datum = inDate;
                    SELECT 1 INTO f_1 FROM feiertage where datum = (inDate + interval '1' day);
                    SELECT 1 INTO f_2 FROM feiertage where datum = (inDate + interval '2' day);
                    SELECT 1 INTO f_3 FROM feiertage where datum = (inDate + interval '3' day);
                    SELECT 1 INTO f_5 FROM feiertage where datum = (inDate + interval '5' day);

   RETURN QUERY SELECT inDate as eingabeDatum, coalesce(f_heute, 0) as f_heute, coalesce(f_1,0) as f_1, coalesce(f_2,0) as f_2, coalesce(f_3,0) as f_3, coalesce(f_5,0) as f_5 ;
            END;
    $$ LANGUAGE plpgsql;

调用该函数仅返回一列以“,”分隔的值:

psql (9.5.12)
Type "help" for help.

tarec=> select feiertag('2017-01-01');
        feiertag        
------------------------
 (2017-01-01,1,0,0,0,0)
(1 row)

我期望不同的列(每个值对应一个列，因为在函数的开头指定了表)，而不是只有一个包含所有值。有谁知道为什么会这样以及我该如何解决这个问题？

谢谢 蒂莫

Answer 1

使用

SELECT *
       FROM feiertag('2017-01-01');

代替

SELECT feiertag('2017-01-01');

以表格形式获取结果。

(将函数视为表格。)