How do I unlink a device from an existing Apple ID. User forgot the username and password of the Apple ID
Hi,
A friend of mine bought a brand new iPad. He manage to enroll it to an Apple ID. However, he forgot the password of the Apple ID account and the device is linked to it. I tried to configure it myself. As i go through the configuration, it says "...This iPad is currently linked to an Apple ID (XXXXX)"
I asked my friend, he forgot the username and password.
Thanks....
Hi
I manage to login to his account and reset the password of his icloud account.
However, now when i activate the ipad using this apple id and the password. It says...
"xxxxxx" cannot be used to unlock this iPad.
I have provided the right Apple ID and Password - still didnt work?
Thanks.
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Hi,
I do want to set the username and password dynamically in hibernate.cfg.xml
Here below is my configuration file.
{code<hibernate-configuration>
<session-factory>
<property name="hibernate.bytecode.use_reflection_optimizer">false</property>
<property name="hibernate.connection.driver_class">oracle.jdbc.driver.OracleDriver</property>
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public class Sessions { private static SessionFactory sessionFactory; private static Configuration configuration = new Configuration(); static { try { String userName = GuigenserviceImpl.userName; String password = GuigenserviceImpl.password; String hostName = GuigenserviceImpl.hostName; String portNo = GuigenserviceImpl.portNo; String sId = GuigenserviceImpl.sId; configuration .setProperty("hibernate.connection.username", userName); configuration .setProperty("hibernate.connection.password", password); configuration.setProperty("hibernate.connection.url", "jdbc:oracle:thin:@" + hostName + ":" + portNo + ":" + sId); try { configuration.configure("/hibernate.cfg.xml"); sessionFactory = configuration.buildSessionFactory(); GuigenserviceImpl.strAccpted = "true"; } catch (Exception e) { e.printStackTrace(); GuigenserviceImpl.strAccpted = "false"; } } catch (HibernateException hibernateException) { GuigenserviceImpl.strAccpted = "false"; hibernateException.printStackTrace(); } catch (Throwable ex) { GuigenserviceImpl.strAccpted = "false"; ex.printStackTrace(); throw new ExceptionInInitializerError(ex); } } public static SessionFactory getSessionFactory() { return sessionFactory; }
So, in this above scenario, suppose if im giving the wrong password means exception should be caught and the string variable "GuigenserviceImpl.strAccpted" should be assigned to false.
But im getting the SQL Exception only in console like wrong username and password. And finally i couldn't able to catch the exception in Sessions class, static block.
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Im getting this following exception message in my console.
INFO: configuring from resource: /hibernate.cfg.xml Apr 6, 2009 2:47:00 PM org.hibernate.cfg.Configuration getConfigurationInputStream INFO: Configuration resource: /hibernate.cfg.xml Apr 6, 2009 2:47:00 PM org.hibernate.cfg.Configuration doConfigure INFO: Configured SessionFactory: null Apr 6, 2009 2:47:00 PM org.hibernate.connection.DriverManagerConnectionProvider configure INFO: Using Hibernate built-in connection pool (not for production use!) Apr 6, 2009 2:47:00 PM org.hibernate.connection.DriverManagerConnectionProvider configure INFO: Hibernate connection pool size: 10 Apr 6, 2009 2:47:00 PM org.hibernate.connection.DriverManagerConnectionProvider configure INFO: autocommit mode: false Apr 6, 2009 2:47:00 PM org.hibernate.connection.DriverManagerConnectionProvider configure INFO: using driver: oracle.jdbc.driver.OracleDriver at URL: jdbc:oracle:thin:@192.168.1.12:1521:orcl Apr 6, 2009 2:47:00 PM org.hibernate.connection.DriverManagerConnectionProvider configure INFO: connection properties: {user=scott, password=****} Apr 6, 2009 2:47:01 PM org.hibernate.cfg.SettingsFactory buildSettings WARNING: Could not obtain connection metadata java.sql.SQLException: ORA-01017: invalid username/password; logon denied at oracle.jdbc.driver.SQLStateMapping.newSQLException(SQLStateMapping.java:70) at oracle.jdbc.driver.DatabaseError.newSQLException(DatabaseError.java:131) at oracle.jdbc.driver.DatabaseError.throwSqlException(DatabaseError.java:204) at oracle.jdbc.driver.T4CTTIoer.processError(T4CTTIoer.java:455) at oracle.jdbc.driver.T4CTTIoer.processError(T4CTTIoer.java:406) at oracle.jdbc.driver.T4CTTIoer.processError(T4CTTIoer.java:399) at oracle.jdbc.driver.T4CTTIoauthenticate.receiveOauth(T4CTTIoauthenticate.java:799) at oracle.jdbc.driver.T4CConnection.logon(T4CConnection.java:368) at oracle.jdbc.driver.PhysicalConnection.<init>(PhysicalConnection.java:508) at oracle.jdbc.driver.T4CConnection.<init>(T4CConnection.java:203) at oracle.jdbc.driver.T4CDriverExtension.getConnection(T4CDriverExtension.java:33) at oracle.jdbc.driver.OracleDriver.connect(OracleDriver.java:510) at java.sql.DriverManager.getConnection(DriverManager.java:525) at java.sql.DriverManager.getConnection(DriverManager.java:140) at org.hibernate.connection.DriverManagerConnectionProvider.getConnection(DriverManagerConnectionProvider.java:110) at org.hibernate.cfg.SettingsFactory.buildSettings(SettingsFactory.java:84) at org.hibernate.cfg.Configuration.buildSettings(Configuration.java:2073) at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1298) at com.beyon.ezygui.server.Sessions.<clinit>(Sessions.java:39) at com.beyon.ezygui.server.GuigenserviceImpl.testRPC(GuigenserviceImpl.java:322) at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39) at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
Thanks in advance.
Thanks & Regards,
Kothandaraman N.Hi,
Myself hardcoded that username and password checking like the below code.
String name = loginData.get("userName").toString();
String pswd = loginData.get("pswd").toString();
String hstName = loginData.get("hName").toString();
String prtNo = loginData.get("portNo").toString();
String sid = loginData.get("sId").toString();
SessionFactory sessionFactory = null;
try {
if (name.trim().equals(userName) && pswd.trim().equals(password)
&& hstName.trim().equals(hostName)
&& prtNo.trim().equals(portNo) && sid.trim().equals(sId)) {
sessionFactory = Sessions.getSessionFactory();
strAccpted = "true";
} else {
strAccpted = "false";
} catch (Exception e) {
e.printStackTrace();
strAccpted = "false";
}I have my own values in string objects, then comparing that values with the values from login form. If both values are matching, then i will do configurations in hibernate.
ResourceBundle resourceBundle = ResourceBundle
.getBundle("com.beyon.ezygui.server.Queries");
public static final String connection_url = resourceBundle
.getString("connection.url");
public static final String userName = resourceBundle.getString("userName");
public static final String password = resourceBundle.getString("password");
public static final String hostName = resourceBundle.getString("hostName");
public static final String portNo = resourceBundle.getString("portNo");
public static final String sId = resourceBundle.getString("sId");The above are the String objects i'm checking for the match with values from login form.
Thanks & Regards,
Kothandaraman N. -
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Dear Apple
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<Edited by Host>This is a user forum. You aren't talking to Apple here.
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<Edited by Host>
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