Question

我需要制作一个PL / SQL类型，只能使用10000000到9999999999999之间的数字（所有正数从8到13个数字）。我正在使用SQL Developer 11g。到目前为止，我已成功使用：

CREATE OR REPLACE TYPE someType AS OBJECT (
tip NUMBER (13)
) FINAL;

CREATE TABLE someTable(
    ID NUMBER PRIMARY KEY,
    PIN someType,
    check ( PIN.tip between 100000000 and 9999999999999)
    )

如何将someTable的check子句放入someType的主体？

CREATE OR REPLACE TYPE someType AS OBJECT (
    tip NUMBER (13)
    ) FINAL;

CREATE TYPE BODY someType as...?

Answer 1

您可以创建CONSTRUCTOR，为超出范围的值引发用户定义的EXCEPTION。

 CREATE OR REPLACE TYPE sometype AS object
(
  tip NUMBER (13),
  CONSTRUCTOR
  FUNCTION sometype ( tip NUMBER )
    RETURN SELF AS RESULT
 ) FINAL; 


CREATE 
OR REPLACE TYPE BODY sometype AS 
 CONSTRUCTOR FUNCTION sometype 
                 ( tip NUMBER )
        RETURN SELF AS RESULT AS tip_beyond_range EXCEPTION;
        PRAGMA EXCEPTION_INIT(tip_beyond_range, -20101);
        BEGIN
                IF tip BETWEEN 100000000
                        AND 9999999999999 THEN
                        self.tip := tip;
                        RETURN;
                  ELSE
                        raise_application_error(-20101, 'ERROR : TIP SHOULD BE BETWEEN 100000000 and 9999999999999');
                END IF;
      END ;
END;

现在，下面的第二个调用将导致错误。

DECLARE
  t1  sometype;

BEGIN
  t1 := NEW sometype(100000000);
  t1 := NEW sometype(10000000);
END;
/    

ORA-20101: ERROR : TIP SHOULD BE BETWEEN 100000000 and 9999999999999 ORA-06512: at "SQL_ZVRGRVBJDGVKVLZDKPHRMMRDP.SOMETYPE", line 12
ORA-06512: at line 6
ORA-06512: at "SYS.DBMS_SQL", line 1721

将约束添加到不同类型PL / SQL的主体

1 个答案: