我正在使用oracle 19 c.我运行一个删除与功能在另一个架构，我得到错误

create user USER1  identified by USER1;
create user USER2  identified by USER2;
create user USER3  identified by USER3;
grant connect to USER1;
grant connect to USER2;
grant connect to USER3;
grant unlimited tablespace to USER1;
grant unlimited tablespace to USER2;
grant unlimited tablespace to USER3;

grant create procedure to USER1;
grant create table to USER1;
grant create view to USER2;

CREATE OR REPLACE PACKAGE USER1.PKG1 IS
  FUNCTION FUNC1(p_DTE DATE) RETURN VARCHAR2;
END PKG1;
/
CREATE OR REPLACE PACKAGE BODY USER1.PKG1 IS
  FUNCTION FUNC1(p_DTE DATE) RETURN VARCHAR2 IS
  BEGIN
      RETURN NULL;
  END;
END PKG1;
/
GRANT ALL     ON USER1.PKG1 TO USER2 WITH GRANT OPTION;
GRANT EXECUTE ON USER1.PKG1 TO USER3;
create table USER1.TAB1(  str1 varchar2(100),  num1 number,  dte1 DATE)
/
insert into USER1.TAB1 (STR1, NUM1, DTE1)values ('S1', 1, SYSDATE);
insert into USER1.TAB1 (STR1, NUM1, DTE1)values ('S2', 2, SYSDATE-1);
COMMIT;
/
GRANT ALL ON USER1.TAB1 TO USER2 WITH GRANT OPTION;
/
CREATE OR REPLACE VIEW USER2.VIW1 AS SELECT * FROM USER1.TAB1
/
GRANT SELECT,INSERT,UPDATE,DELETE ON USER2.VIW1 TO USER3;

字符串
后创建环境
使用低于代码的用户3登录时出现错误：ORA-01732：此视图上的数据操作不法律的

DELETE USER2.VIW1 WHERE USER1.PKG1.FUNC1(DTE1)='0';

型
使用以下代码的用户2登录成功运行。

DELETE USER2.VIW1 WHERE USER1.PKG1.FUNC1(DTE1)='0';

型