postgresql - 使用函数创建具有基本内部连接的简单选择

Question

我是 postgresql 的新手，我已经掌握了 join 语句的基本知识，现在我正试图从中创建一个函数

select r.studid,r.subjcode,r.section,s.subjdesc,r.sy,r.sem,r.grade,r.gcompl
from registration r
inner join subject s ON r.subjcode = s.subjcode
where studid = '2020-2001' AND r.grade = 'INC' AND COALESCE(gcompl, '') = ''
ORDER BY studid DESC;

我目前正在使用带有 pgadmin III 编辑的 postgresql v9.5 我得到了它的工作

CREATE OR REPLACE FUNCTION srgb.viewincgrade(p_studid char)
RETURNS TABLE (studid char,subjcode char,section char ,subjdesc 
varchar,sy char,sem char,grade varchar,gcompl varchar)
AS
$$
BEGIN
return query
select 
r.studid,r.subjcode,r.section,s.subjdesc,r.sy,r.sem,r.grade,r.gcompl
from registration r
INNER JOIN subject s ON r.subjcode = s.subjcode
where r.studid = p_studid AND r.grade = 'INC' AND COALESCE(r.gcompl, 
'') = ''
ORDER BY studid DESC;
END
$$
LANGUAGE plpgsql;