Passing the username and password via report link

How can I pass the username and password via the reprt link?
thanks all.
Ahmad Esbita

Hi,
Jigar Oza
Thanks for u r reply, i have tried with application integrator but there is a problem with usermapping.
now what i have done is created HTTP System based on this system created URL iView,in application parameters username and password as MappedUser and MappedPassword every thing is working fine. user logged in automatically when he logged into portal.
there are tabs ,links in application .when i click on tabs or links it is assking to enter username and password of the application.
did i do any thing wrong in creating HTTP system or URL iView,
what are the necessary properties should be given.
replys are highly appreciated.
Regards
K Naveen Kishore

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Dynamically pass the username and password to report server to invoke reports

Hi
The scenario is as follows:-
9iAS is installed under - Windows 2000 (web server and report Server 6i) and everything works fine.
We have developed a page using pl/sql cartrige (oracle 8i), from that a report is invoked with some parameters. However, the userid and password is stored under keymap file (cgicmd.dat).
Is there anyway to pass dynamically the username and password to report server in order to invoke the report.?
Can any one advise?
null

Check out
http://serverfault.com/questions/371907/can-you-pass-user-pass-for-http-basic-authentication-in-url-parameters
Kind regards,
Margriet Bruggeman
Lois & Clark IT Services
web site: http://www.loisandclark.eu
blog: http://www.sharepointdragons.com

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how to Pass the username and password in URL to auto log in the application from SharePoint List
I have a link in my SharePoint list when user try to open the link its asking username and password .i want to put username and password in URL to auto log in into that external URL from sharepoint
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Check out
http://serverfault.com/questions/371907/can-you-pass-user-pass-for-http-basic-authentication-in-url-parameters
Kind regards,
Margriet Bruggeman
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Hi Gurus,
i Have Created URL iView. i want to send the Username and password of portal to this url ,
how can i access the portal username and password.
Note: tried with j_user and j_password
Regards
K Naveen Kishore

Hi,
Jigar Oza
Thanks for u r reply, i have tried with application integrator but there is a problem with usermapping.
now what i have done is created HTTP System based on this system created URL iView,in application parameters username and password as MappedUser and MappedPassword every thing is working fine. user logged in automatically when he logged into portal.
there are tabs ,links in application .when i click on tabs or links it is assking to enter username and password of the application.
did i do any thing wrong in creating HTTP system or URL iView,
what are the necessary properties should be given.
replys are highly appreciated.
Regards
K Naveen Kishore

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If your friend refuses to give your money back and can not or will not remove the activation lock, there is a fair possibility that the phone was not theirs to sell in the first place.

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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>
       <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>
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</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.

Can't get FaceTime to work...any suggestions...it just keeps me in this loop of asking to sign in, then verifying, then asking to sign in again. I know the username and password are correct.

Can't get face time to work. Know the username and password are right. Asks me to sign in...verifying... And then goes back to sign in...

One thing that worked for me:
launched Message app which uses the same backend as Facetime. If it launches, click on the video. May display a Error Dialog. Call AppleCare and give them the number displayed. This will give them the needed info to check your AppleID account for the messaging servers
For some reason FaceTime did not display this error, yet once I gave the Support person the number it was fixed in just a minute.
YMMV

Dear Apple I buy a refurbish iPhone 5s the IMEI ******* and ICCID ******** and i restored the device when i restored its locked to iCloud and i don't know the username and password i hope to help me about the problem asap. Best regard

Dear Apple
I buy a refurbish iPhone 5s the IMEI ******** and ICCID ********** and i restored the device when i restored its locked to iCloud and i don't know the username and password i hope to help me about the problem asap.
Best regard
<Edited by Host>

This is a user forum. You aren't talking to Apple here.
Return your iPhone for a refund. There is no way to remove the Activation Lock. Apple won't get involved.
<Edited by Host>

Passing the username and password via report link

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