Java: Detecting if a variable is a String or an Integer

Question

I'm look for some help on a bit of homework I have. I want the user to enter a numerical String, then convert it to an Integer. But I want to make a loop that will detect if the user entered in the wrong value such as "One Hundred" as apposed to "100".

What I was thinking was to do something like this:

    do{
        numStr = JOptionPane.showInputDialog("Please enter a year in numarical form:"
                        + "\n(Ex. 1995):");
        num = Integer.parseInt(numStr);
            if(num!=Integer){
            tryagainstr=JOptionPane.showInputDialog("Entered value is not acceptable."
                                  + "\nPress 1 to try again or Press 2 to exit.");
    tryagain=Integer.parseInt(tryagainstr);
            }
            else{
            *Rest of the code...*
            }
            }while (tryagain==1);

But I don't know how to define that "Integer" value. I essentially want it to see if it is a number or not to prevent it from crashing if the user enters the wrong thing.

Integer.parseInt method throws NumberFormatException if the input is not parseable as an integer. You just have to use a try/catch. — Bhesh Gurung
– Bhesh Gurung, Commented Feb 19, 2013 at 16:07

Mina Fawzy · Accepted Answer · 2015-02-23 09:59:39Z

9

Try use instanceof , this method will help you with check between many types

Example

if (s instanceof String ){
// s is String
}else if(s instanceof Integer){
// s is Integer value
}

If you want just check between integer and string you can use @NKukhar code

try{
        Integer.valueOf(str);
    } catch (NumberFormatException e) {
        //not an integer
    }

answered Feb 23, 2015 at 9:59

Mina Fawzy

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Comments

nkukhar · Accepted Answer · 2013-02-19 16:11:19Z

5

Try this:

    try{
        Integer.valueOf(str);
    } catch (NumberFormatException e) {
        //not an integer
    }

answered Feb 19, 2013 at 16:11

nkukhar

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Comments

Jashaszun · Accepted Answer · 2014-08-06 22:13:23Z

2

try this

int num;
String s = JOptionPane.showInputDialog("Enter a number please");
while(true)
{
    if(s==null) 
        break; // if you press cancel it will exit
    try {
        num=Integer.parseInt(s);
        break;
    } catch(NumberFormatException ex)
    {
        s = JOptionPane.showInputDialog("Not a number , Try Again");
    }
}

edited Aug 6, 2014 at 22:13

Jashaszun

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answered Feb 19, 2013 at 21:05

Azad

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1 Comment

Dave555 Over a year ago

Thank you! I understand the try/catch better now. I ran something similar to this where I told it to do{ try/catch }while a variable equaled 1, and every time it went to the catch the variable stayed 1 to keep looping. Thanks!

Fritz · Accepted Answer · 2013-02-19 16:23:23Z

1

Use a regex to validate the format of the string, and accept only numeric values on it:

Pattern.matches("/^\d+$/", numStr)

The matches method will return true if numString contains a valid numeric sequence, but of course the input can be way above an Integer's capacity. In that case, you can consider switching to a long or a BigInteger type.

answered Feb 19, 2013 at 16:23

Fritz

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