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In Topic: Java Session objects

Posted 28 Nov 2011

Yes, thank you so much .now i got it ..

pbl, on 28 November 2011 - 10:30 PM, said:

Create a Permission class that contains the permissions, or an int with flags
Associate a Permission object, or a int, to all your users
Pass it as parameter to your new UserPage(); constructor

String[] username = {"Peter", "Paul", "Ringo", "John"};
String[] password = {"mac", "mosty, "g00se", "fuzzy"};
int[] permission = {1, 2, 3, 4};

for(int i = 0; i < username.length; i++) {
    if(username[i].equalsIgnoreCase(name1) {
       if(password[i].equalsIgnoreCase(pw1) {
          UserPage up = new UserPage(permisssion[i]);
          up.show();
          return;
       else {
         ... invalid password
       }
       break;
     }
}
... invalid username

In Topic: Java Session objects

Posted 28 Nov 2011

Its not multithread program... its simple ..even there will be 3/4 hard coded users.
the main focus of this program is to make session object and limit the access. after login only admin user( user = admin, password= admin) have access to click add user and delete user button .. even i dont need to implement the add/delete feature. If admin user click on add user button i have to display " new user create message" . if other user click on this button the message will be " You have no right to add user " .. but i have to maintain it by making session objects.... hoping for your suggestion.. plz check ur message i send me email.

pbl, on 28 November 2011 - 10:03 PM, said:
Yes but is it multithread ?
How does your users start the application ?
Do 20 users around the world share the same GUI in the same host ?

Its not multithread program... its simple ..even there will be 3/4 hard coded users.
the main focus of this program is to make session object and limit the access. after login only admin user( user = admin, password= admin) have access to click add user and delete user button .. even i dont need to implement the add/delete feature. If admin user click on add user button i have to display " new user create message" . if other user click on this button the message will be " You have no right to add user " .. but i have to maintain it by making session objects.... hoping for your suggestion.. plz check ur message i send me email.

pbl, on 28 November 2011 - 10:03 PM, said:
Yes but is it multithread ?
How does your users start the application ?
Do 20 users around the world share the same GUI in the same host ?
In Topic: Java Session objects

Posted 28 Nov 2011

Yes, PBL . it is GUI program and it is my assignment so i cant copy and pest whole my code here.. can i send you my whole program code in ur email/message?

pbl, on 28 November 2011 - 09:43 PM, said:
Are you talking of a multi thread application/GUI where users connect as Client and share the same GUI ?
How the users start your application ?
In Topic: Java Session objects

Posted 28 Nov 2011
yes i did mistake in typing
```
UserPanel obj= new UserPanel();
obj.showform();
```
userpage is another class that contain the information after userlogin. different user have different information to acess.

pbl, on 28 November 2011 - 09:20 PM, said:
First,m this won't compile
```
Obj=new UserPage();
Obj.showform();
```
unless OBJ is defined somewhere else as a UserPage()

And what is a UserPage ?

What prompts the user for username and password, that what is calling Login ?
bhojnp, on 28 November 2011 - 09:28 PM, said:
yes i did mistake in typing
```
UserPanel obj= new UserPanel();
obj.showform();
```
userpage is another class that contain the information after userlogin. different user have different information to acess.

pbl, on 28 November 2011 - 09:20 PM, said:
First,m this won't compile

Obj=new UserPage(); Obj.showform();

unless OBJ is defined somewhere else as a UserPage()

And what is a UserPage ?

What prompts the user for username and password, that what is calling Login ?

In Topic: Java Syntax

Posted 22 Nov 2011

thanks for all of your comments now i understand little bit . actually i tried to check below case and found the wrong answer when i was used ==

String original1= " hello";
String reverse= "olleh";

//Case 1
if (original1.equals(reverse))
System.out.println("both are equal");
else
System.out.println("both are not equal";

//case 2
if (original1==reverse))
System.out.println("both are equal");
else
System.out.println("both are not equal";

for the case 2 i found answer for palindrome check

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