1. 后台批量更新用户密码
DECLARE
l_success BOOLEAN;
BEGIN
dbms_output.enable(300000);
FOR i IN (SELECT fu.user_name
FROM fnd_user fu
WHERE fu.creation_date > to_date('2013-01-01', 'yyyy-mm-dd')) LOOP
l_success := fnd_user_pkg.changepassword(i.user_name, '111111');
IF l_success THEN
dbms_output.put_line(i.user_name || '-' || '密码已更新');
ELSE
dbms_output.put_line(i.user_name || '-' || '密码更新失败');
END IF;
END LOOP;
END;
2. 用户密码解密
首先建个函数
CREATE OR REPLACE Function Cux_Fnd_Decrypt(Key In Varchar2,
Value In Varchar2)
Return Varchar2 As
Language Java Name 'oracle.apps.fnd.security.WebSessionManagerProc.decrypt(java.lang.String,java.lang.String) return java.lang.String';
然后执行下列SQL:
Select fu.user_id,Fu.User_Name, Cux_Fnd_Decrypt('XXXXX', Fu.Encrypted_User_Password)
From Fnd_User Fu
WHERE fu.user_name='SYSADMIN'
把其中的XXXXX替换成APPS 密码即可.
DECLARE
l_success BOOLEAN;
BEGIN
dbms_output.enable(300000);
FOR i IN (SELECT fu.user_name
FROM fnd_user fu
WHERE fu.creation_date > to_date('2013-01-01', 'yyyy-mm-dd')) LOOP
l_success := fnd_user_pkg.changepassword(i.user_name, '111111');
IF l_success THEN
dbms_output.put_line(i.user_name || '-' || '密码已更新');
ELSE
dbms_output.put_line(i.user_name || '-' || '密码更新失败');
END IF;
END LOOP;
END;
2. 用户密码解密
首先建个函数
CREATE OR REPLACE Function Cux_Fnd_Decrypt(Key In Varchar2,
Value In Varchar2)
Return Varchar2 As
Language Java Name 'oracle.apps.fnd.security.WebSessionManagerProc.decrypt(java.lang.String,java.lang.String) return java.lang.String';
然后执行下列SQL:
Select fu.user_id,Fu.User_Name, Cux_Fnd_Decrypt('XXXXX', Fu.Encrypted_User_Password)
From Fnd_User Fu
WHERE fu.user_name='SYSADMIN'
把其中的XXXXX替换成APPS 密码即可.
来自 “ ITPUB博客 ” ，链接：http://blog.itpub.net/10359218/viewspace-1063348/，如需转载，请注明出处，否则将追究法律责任。
请登录后发表评论
登录
全部评论
