Question

我在创建返回多行的函数时遇到问题。我试图使用一个循环，但我仍然得到“oracle精确提取返回超过请求”错误。任何指导我正确方向的帮助将不胜感激。

CREATE OR REPLACE FUNCTION sections2
(
id_param NUMBER
)
RETURN VARCHAR
AS
sections_param VARCHAR(40);
BEGIN
LOOP
SELECT sections
INTO sections_param
FROM table2
WHERE id = id_param;
RETURN sections_param;
end loop;
END;
/

Answer 1

听起来你想要一个返回集合的函数。我将展示一个返回嵌套表的示例

-- Use whatever length is appropriate here.  
CREATE TYPE sections_tbl
    IS TABLE OF VARCHAR2(40); 

CREATE OR REPLACE FUNCTION sections2 (
  id_param NUMBER
)
  RETURN sections_tbl
AS
  l_sections sections_tbl;
BEGIN
  SELECT sections
    BULK COLLECT INTO l_sections
    FROM table2
   WHERE id = id_param;
  RETURN l_sections;
END;

在来电者中，您可以执行类似

的操作

DECLARE
  l_sections sections_tbl := sections2( <<id>> );
BEGIN
  FOR i IN 1 .. l_sections.count
  LOOP
    dbms_output.put_line( l_sections(i) );
  END LOOP;
END;

oracle exact fetch返回超过请求的数量

1 个答案: