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java very basic input problem

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Join Date: Apr 2008

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localp

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java very basic input problem

Nov 15th, 2008

i want to pass an integer value as an input, for that first i have to get the value as an string and then convert it into an Integer. i wrote the code and it looked like this ..

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
String age="";
int intage=0;
       System.out.print("Please Enter your age : ");
 
        try{
            age= dataIn.readLine();
            intage=Integer.parseInt(age);
        }catch( IOException e ){
            e.printStackTrace();
        }
String age="";
int intage=0;
       System.out.print("Please Enter your age : ");
               
        try{
            age= dataIn.readLine();
            intage=Integer.parseInt(age);
        }catch( IOException e ){
            e.printStackTrace();
        }

but my problem here is that when i enter a String like "dfjdkdjre" , as input , i get an error or rather an exeption.. how do i prevent this ... can i display an error message or ??

can some one help me complete my code .. please

and could some one tell me what does this line do...

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
e.printStackTrace();
e.printStackTrace();

please help me .. i am kind of lost in this

Local P ...

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jasimp

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Re: java very basic input problem

Nov 15th, 2008

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
e.printStackTrace()
e.printStackTrace()

is used for "tracing" where your exception occurred. It is useful to the programmer for debugging, but useless to the end user. Try validating your input before you parse it to an integer. If you catch that exception you get from

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
intage=Integer.parseInt(age);
intage=Integer.parseInt(age);

you can handle it accordingly.

Last edited by jasimp; Nov 15th, 2008 at 2:00 pm.

"Argyou not with the hand you are dealt in cards or life." ---- Wizard and Glass

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Join Date: Apr 2008

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Reputation: localp is an unknown quantity at this point

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localp

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Re: java very basic input problem

Nov 15th, 2008

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Originally Posted by jasimp

Help with Code Tags

Java Syntax (Toggle Plain Text)
e.printStackTrace()
e.printStackTrace()
is used for "tracing" where your exception occurred. It is useful to the programmer for debugging, but useless to the end user. Try validating your input before you parse it to an integer. If you catch that exception you get from
Help with Code Tags

Java Syntax (Toggle Plain Text)
intage=Integer.parseInt(age);
intage=Integer.parseInt(age);
you can handle it accordingly.

it didnt work, i get an error when i type a string like " hhfdh "; it works for numbers but not for string ...

Local P ...

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VernonDozier VernonDozier is offline

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Re: java very basic input problem

Nov 15th, 2008

•

Originally Posted by localp

it didnt work, i get an error when i type a string like " hhfdh "; it works for numbers but not for string ...

I don't know what you did, but you perhaps aren't catching a NumberFormatException. IOException won't catch the error you are referring to. parseInt throws a NumberFormatException in the situation you are referring to. If you are only catching IOException, you are going to get an error.
http://java.sun.com/j2se/1.5.0/docs/...va.lang.String)

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tonief

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Re: java very basic input problem

-1

Nov 16th, 2008

You can try like this

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
Buffered reader tast=new BufferedReader( new InputStreamreader(System.in));
     String age="";
     age=tast.readLine();
     intage=Integer.parseInt(age);
Buffered reader tast=new BufferedReader( new InputStreamreader(System.in));
     String age="";
     age=tast.readLine();
     intage=Integer.parseInt(age);

Last edited by tonief; Nov 16th, 2008 at 8:10 am.

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jasimp

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Re: java very basic input problem

Nov 16th, 2008

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Originally Posted by tonief

You can try like this
Help with Code Tags

Java Syntax (Toggle Plain Text)
Buffered reader tast=new BufferedReader( new InputStreamreader(System.in));
     String age="";
     age=tast.readLine();
     intage=Integer.parseInt(age);
Buffered reader tast=new BufferedReader( new InputStreamreader(System.in)); String age=""; age=tast.readLine(); intage=Integer.parseInt(age);

They already have that. If you actually read the thread they were having trouble with an exception.

"Argyou not with the hand you are dealt in cards or life." ---- Wizard and Glass

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stultuske

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Re: java very basic input problem

Nov 17th, 2008

•

Originally Posted by localp

but my problem here is that when i enter a String like "dfjdkdjre" , as input , i get an error or rather an exeption.. how do i prevent this ... can i display an error message or ??

by making sure to use only valid input..
or, just by replacing

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
e.printStackTrace();
e.printStackTrace();

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
System.out.println(age + " is not a valid Integer.");
System.out.println(age + " is not a valid Integer.");