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Nov 15th, 2008

java very basic input problem

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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 ..

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...

Java Syntax (Toggle Plain Text)
e.printStackTrace();
e.printStackTrace();

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

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localp

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Nov 15th, 2008

Re: java very basic input problem

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

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.

jasimp

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Nov 15th, 2008

Re: java very basic input problem

Click to Expand / Collapse Quote originally posted by jasimp ...

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
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 ...

localp

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Nov 15th, 2008

Re: java very basic input problem

Click to Expand / Collapse Quote 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)

VernonDozier

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Nov 16th, 2008

-1

Re: java very basic input problem

You can try like this

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.

tonief

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Nov 16th, 2008

Re: java very basic input problem

Click to Expand / Collapse Quote originally posted by tonief ...

You can try like this
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.

jasimp

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Nov 17th, 2008

Re: java very basic input problem

Click to Expand / Collapse Quote 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

Java Syntax (Toggle Plain Text)
e.printStackTrace();
e.printStackTrace();

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

stultuske

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