創建oracle函數

Question

以下代碼使用oracle sql * plus創建一個包含get_age函數的對象表。

Create Type C_type as Object(
    Cname varchar(30), 
    dob Date, 
    classification Number, 
    sample_clip Blob 
) 
/

ALTER TYPE C_type 
    ADD MEMBER FUNCTION get_age RETURN INTEGER; 


CREATE OR REPLACE TYPE BODY C_type AS 
MEMBER FUNCTION get_age RETURN INTEGER IS 
currenttime_age integer; 
BEGIN 
    currenttime_age := (SYSDATE - dob)/365.25; 
    RETURN currenttime_age; 
END get_age; 
END; 
/

我打算是添加返回字符

• 分類爲「短」（如果低於5）的另一功能，「中」（5-45之間）和「長」（以上45）

Answer 1

試試這個。我的錯。

DROP TYPE c_type 
/

CREATE TYPE C_type AS OBJECT 
(
    Cname VARCHAR (30), 
    dob DATE, 
    classification NUMBER, 
    sample_clip BLOB, 
    MEMBER FUNCTION get_age 
     RETURN INTEGER, 
    MEMBER FUNCTION get_class 
     RETURN VARCHAR2 
) 
/


CREATE OR REPLACE TYPE BODY C_type 
AS 
    MEMBER FUNCTION get_age 
     RETURN INTEGER 
    IS 
     currenttime_age INTEGER; 
    BEGIN 
     currenttime_age := (SYSDATE - dob)/365.25; 
     RETURN currenttime_age; 
    END get_age; 


    MEMBER FUNCTION get_class 
     RETURN VARCHAR2 
    IS 
     result VARCHAR2 (10); 
    BEGIN 
     result := 
     CASE 
      WHEN classification < 5 THEN 'short' 
      WHEN classification BETWEEN 5 AND 45 THEN 'medium' 
      WHEN classification > 45 THEN 'long' 
     END; 

     RETURN result; 
    END get_class; 
END; 
/