Try and catch Help!

[Beth Laguardia]

Hello guys I am trying to play with try and catch but i cannot get the results that I want. Hope you will help me out.

Here's the program:

  import java.io.*;
public class UpChar
{
  public static void main(String[] args) throws IOException
  { String name= ""s;
    int i=0;
    BufferedReader dataIn = new BufferedReader(new InputStreamReader(System.in));
   
    try {
 
    System.out.println("Enter your name: ");
    name=dataIn.readLine();
    
    
     }
 
   catch (NumberFormatException e) {
    System.out.println("Error!");
    i=1;
    
   }
  
   
    if (i==0)
    System.out.println("Hello, " +name.toUpperCase() + "How are you doing?");
   
    
  }
}  

What I want to do is catch and print error if the user input an integer. But in the progam even thou i input an integer it will still pint it. Can you please help me? I have not taken his topic before because i am a very much newbie with java.

Thanks

[Ernest Friedman-Hill]

Hi Beth,

Nothing in your program will actually throw a NumberFormatException, so you won't ever catch one. NumberFormatException is thrown by methods that want a string to contain an integer, if it turns out that the string doesn't. For example, the "Integer.valueOf()" will throw an exception here, and the catch block will execute:

try {
    Integer fail = Integer.valueOf("nothing like an integer");
} catch (NumberFormatException nfe) {
    System.out.println("Ooops!");
}

[Matt Swaggi]

hai friend ! i hope this coding works ! it keeps on asking the user to give name as long as he types the character ! coding is asfollows

***************************************************************************

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
 
 
public class Ride {
int s ;
String d ;
void getdata()
{
try
{
do
{
System.out.println("Enter the NAME " );
 BufferedReader br =  new BufferedReader(new InputStreamReader(System.in));
       d = br.readLine(); 
 }
while(!d.matches("[a-z,A-Z]"));
}
 catch(Exception e)
 {
      
 }

  
 
}

public static void main(String[] args) {
// TODO Auto-generated method stub
Ride e = new Ride();
e.getdata();
 
}
 
}

[edit]Add code tags. CR[/edit]
[ June 15, 2008: Message edited by: Campbell Ritchie ]

[Beth Laguardia]

Originally posted by Ernest Friedman-Hill:
Hi Beth,

Nothing in your program will actually throw a NumberFormatException, so you won't ever catch one. NumberFormatException is thrown by methods that want a string to contain an integer, if it turns out that the string doesn't. For example, the "Integer.valueOf()" will throw an exception here, and the catch block will execute:

try {
    Integer fail = Integer.valueOf("nothing like an integer");
} catch (NumberFormatException nfe) {
    System.out.println("Ooops!");
}

Hello ma'am,

May I know what should i do so my program can catch integer entered? What i want my program to do is accept string not integer but it still accept integers. Is there a way? Thanks

[Rob Spoor]

Originally posted by Beth Laguardia:
Hello ma'am,

May I know what should i do so my program can catch integer entered? What i want my program to do is accept string not integer but it still accept integers. Is there a way? Thanks

Surely you're not calling Ernest ma'am?? Dr. Friedman-Hill is quite male indeed

If you want to disallow numbers then you should do it the other way around: try parsing the name into an integer, and if that fails then it's a GOOD thing:

try
{
    System.out.println("Enter your name: ");
    name = dataIn.readLine();
    i = Integer.parseInt(name);
    // if the following line will run then name was a valid int
    System.out.println("Error!");
}
catch (NumberFormatException e)
{
    // name is not a valid int - that's a good thing here
    // do stuff
}

[Matt Swaggi]

Hey
Beth Laguardia wont my coding is not working for You ! i hope you may not run the coding ! well anyway i said my suggestion its up to You !
bye !!

[Ulf Dittmer]

If you want to make sure it's not an integer you would quit if you were successful parsing the string into an integer. Something like this:

try { 
    System.out.println("Enter your name: ");
    name=dataIn.readLine();
    Integer.parseInt(name);
    // we got here without an exception, so it IS an integer - bye bye!
    System.exit(1);
} catch (NumberFormatException e) {
    // it's not an integer, so we can ignore the exception
    // (it may of course not be a proper name either)
}

There are more graceful ways of writing this code (using string operations or regular expressions), but for starters this may suffice.

[Beth Laguardia]

Originally posted by GUNA RANJAN:
Hey
Beth Laguardia wont my coding is not working for You ! i hope you may not run the coding ! well anyway i said my suggestion its up to You !
bye !!

Hey Gina I am sorry. I thought the code you posted is yours and you are having a hard time also. Will try all the sugestion here. Its the only way I can learn..

Thanks a lot

[Beth Laguardia]

Thank you all sir!

Will try all the codes and study it. Really thanks a bunch!