Logging in Enterprise Manager by wrong username and password?

Kindly, some one explain that why I am able to login in to Enterprise Manager (stand Alone - Mode) by any wrong username and password as sysdba privilege.

If you have the privilege to connect without userid/password (i.e. connect / as sysdba works), userid/password is irrelevant (because you don't need to know neither a valid user nor a valid password, Oracle says, you may type in what you want, nothing or anything else).

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HT204053 I cannot log into iCloud using my Apple username and password. I keep getting an error message that states "CANNOT SIGN UP - The Apple ID is valid but is not an iCloud account." How do I fix this?

I cannot log into iCloud using my Apple username and password. I keep getting an error message that states "CANNOT SIGN UP - The Apple ID is valid but is not an iCloud account." How do I fix this?

You are getting this message because you are attempting to create an iCloud account on a PC. You can only create iCloud account on an iOS device (iPhone, iPad or iPod Touch) running iOS 5 or higher, or on a Mac running OS X Lion (10.7.2) or higher. After creating your account on one of these devices you will then be able to sign into the account on your PC.

Why do I have to type my username and password each time I log in?(My husband's username and password are always in a dropdown box)

When I go to the wild blue log in and click on the username box there is a blue drop down and it shows my husband's user name.
When I click on it my husband's username and password are automatically entered and all I have to do is click on the login box.
But I have to type in my username and password before I can log
in. I want my username and password in the blue drop down box along with my husband's. Then I won't have to type it in each time.

Are you referring to the SAME computer/login?
If not, he likely has auto-complete (or similar) enabled and you probably do not.

Why does my Apple TV tell me I am signing my wrong username and password when I have not

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Can anyone please help me before I explode..:-)
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I tried creating a Linux VM (Cent OS based) using PowerShell. I was able to provision the VM, but when i tried to log in using the credentials in the VM using Putty, I get an error message stating "No supported authentication methods available
(server sent : publicKey).
Although when i create a VM through Azure portal using the same image i don't find any issues to login the VM. Please find below the PowerShell commands i used to provision Linux VM.
Option 1:
New-AzureQuickVM -Linux -ServiceName "customdatalinux" -Name "customdatalinux" -ImageName "0b11de9248dd4d87b1862
1318e037d37__RightImage-CentOS-7.0-x64-v14.2" -Password "******" -Location "East US" -LinuxUser "*****" -InstanceSize "ExtraSmall"
Option 2:
PS C:\> $vm1 = New-AzureVMConfig -Name "customdatalinux" -InstanceSize "ExtraSmall" -ImageName "0b11de9248dd4d87b18621318e037d37__RightImage-CentOS-7.0-x64-v14.2" | Add-AzureProvisioningConfig -Linux -LinuxUser "*****"
-Password "*****"
New-AzureVM -ServiceName "customdatalinux" -VMs $vm1
Option 3:
$vm1 = New-AzureVMConfig -Name "customdatalinux" -InstanceSize "ExtraSmall" -ImageName "0b11de9248dd4d87b1862131
8e037d37__RightImage-CentOS-7.0-x64-v14.2" | Add-AzureProvisioningConfig -Linux -LinuxUser "****" -Password "*****" | Set-AzureEndpoint -Name "SSH" -Protocol "TCP" -Loc
alPort "22" -PublicPort "22"
New-AzureVM -ServiceName "customdatalinux" -VMs $vm1
In the Azure portal i can see that there is an option to select credentials authentication alone, where i can enter the password and provision a VM. I try to do the same through PowerShell and VM gets provisioned, but am unable to log in the VM using those
credentials.
Can you please let me know how to resolve this issue?
Thanks,
Anish

Try this
azure vm create customdatalinux0b11de9248dd4d87b18621318e037d37__RightImage-CentOS-7.0-x64-v14.2 username password -l "West US"By default, linux VM on Azure has Port 22 opened so you don't have to configure those unless you want to change port. Also loose | and '' in commands.
Creating new VM creates new service unless you want to use one already created.
Mustafa Toroman

Discover Installed: How to Setup username and password

I Installed Discover 10.1.2 and BI. When I login with
http://localhost:18101/
It is asking username and password.
1) How to setup the users.
2) Any Sample schemas.
(enterprise-manager requires a username and password.)
Message was edited by:
ksridhar69

I got it.
Enterprise Manager Application Server Control
username:ias_admin
password: specified during install

I cant get my iphone 4 tolog in to my company wifi. i updated to ios 6 and now it just goes to my companies login screen. i enter my username and password and it just sticks. however it will logon to wifi without a requirement for a user name and password

I have to log onto my companies wifi by first going thru a log on page that equires a username and password. before i updated to ios 6. it worked fine now it just goes to the login page and when i hit enter it doesnt do anything just stays stuck on that page. however the phone will log onto a regular wifi router that doesnt require any kind of username or password. any ideas on how to fix it. i have tried everything including reseting all network settings.

1. Settings>General>Reset>Reset Network Settings
or
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How to catch exception while validating the username and password in hbm

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>
       <property name="hibernate.connection.pool_size">10</property>
       <property name="show_sql">true</property>
       <property name="hibernate.dialect">org.hibernate.dialect.OracleDialect</property>
       <property name="hibernate.hbm2ddl.auto">update</property>
       <property name="current_session_context_class">thread</property>
       <property name="show_sql">true</property>
     </session-factory>
</hibernate-configuration>{code}
Also im getting that session factory object like the below code.
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.
Anyone can help me in catching that SQL Exception in my Sessions class and i need to handle that SQL Exception in my class.
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.

How to check Username and Password

hai
in my application i'm navigating from Login page to Home Page . in login page i have to check for particular user name and password let us say abc and abc. if the user enters wrong username and password it should redirect to login page. is there any snippet of code

Hi Sravan,
Import this package in your login view
com.sap.tc.webdynpro.progmodel.controller.MessageManager;
And if your context attributes are Name and Password.And if you are navigating from LoginView to HomeView. Write the code given below inside your action event.
String login=wdContext.currentContextElement().getName();
String password=wdContext.currentContextElement().getPassword();
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          wdThis.wdFirePlugToHomeView();
     else
MessageManager msgMgr= (MessageManager) wdThis.wdGetAPI().getComponent().getMessageManager();
msgMgr.reportException("Login failure.Pls check login details",true);
Hope this will help you
Regards,
Rajeev

I couldn't access a website because I used an incorrect username and password.How do I reaccess with correct username and password. Thanks.

Getting into Avenue5Consulting members only area, I mistakenly used my other username and password. A message came up asking if I wanted to save it and I pressed yes. This didn't open the site, but brought me back to login page. Realizing that I logged in with erroneous username and password, I checked and logged in again with the right username and password, but these were not accepted. Please help! I need to get back to the site to finish my homework!
Will greatly appreciate your support.

That issue can be caused by corrupted cookies.
Clear the cache and the cookies from sites that cause problems.
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I have installed oracle 10g in my winxp machine. i am using enterprise manager. i am not able to shutdown the system as it asks for os username and password. i have provided my os username and password but eerror persiste. somebody pls help me to get rid

I have installed oracle 10g in my winxp machine. i am using enterprise manager. i am not able to shutdown the system as it asks for os username and password. i have provided my os username and password but eerror persiste. somebody pls help me to get rid of this.

Here I am using Java Type IV for database
connection.
So,there was no necessity of creating DNS.How your app communicates with db shouldn't matter for the end user. Still, you may want to use a functional network name also for the thin client driver connection string.
So,is there any other way to solve this problem.What is the problem really? Do you not use dns for network naming? Maybe you have to manage the hosts file on every client then.

Browser prompting for username and password to access enterprise manager.

I've just installed Internet Directory 10.1.4.0.1 and am having trouble connecting to Application Server Console in Enterprise Manager.
When I try to go to the url http://localhost:18101, a dialog pops up asking for a username and password. In firefox it says 'Enter username and password for "enterprise-manager" at http://localhost:18101'. IE6 just says, 'enterprise-manager'. I have tried various sets of credentials for this - my network userid and password; and ias_admin, orcladmin, sys and sysman, all with the passwords I specified at install time. None of these work. If I cancel the dialog box I get a '401 Unauthorised' page. I'm sure this is coming from apache and is basic web page authentication, and not an enterprise manager login.
Has any one seen this before?
I can connect through directory manager, and I can connect to the database console for the metadata repository.
emctl status iasconsole shows that it is running.
Windows XP Pro SP2 standalone (ie not part of a windows domain)

trying to access my college intranet form home
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My mail has stopped receiving anything and is saying the IMAP server username and password is wrong but I haven't logged out or changed anything?

My mail (hotmail) has stopped receiving emails and keeps saying the IMAX server username and password is wrong when I haven't changed anything?

Hello ,Marshyy96
That is not good, lets see if we can get your email back up and running again. Take a look at the article below to go over the settings to just make sure nothing is out of the ordinary. You may end up needing to remove the email account and add it back in.
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Re: Invalid username and password when logging onto Hyperion workspace

Hi,
I have installed Hyperion Reporting and Analysis services.When I attempt to log on to Workspace, I get the following message;
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I was told that Dsun.net.inetaddr.ttl=0" is required to be passed to the web application server (JVM) that starts Hyperion Shared Services and all System 9 products that depend on Hyperion Shared Services
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2010-03-24 08:26:03,509 [[ORB=_it_orb_id_1,Pool=1]::id-12] WARN com.hyperion.css.spi.impl.ldap.LDAPProvider.authenticate(Unknown Source) - javax.naming.ServiceUnavailableException: ldaps.lirr.org:636; socket closed; remaining name ''

Mail says my gmail usernames and passwords are wrong but they're not

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Logging in Enterprise Manager by wrong username and password?

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