How to catch an error if user inputs a String instead of Int (JOptionPane)

Question

    int [] numbers = {1,2,3,4};
    Random random = new Random();
    int totalGood=0;
    
    int totalFalse=0;
    String titl = "title1";
    String titl2 = "Question";
    String titl3 = "Error!";
    boolean ages = true;  

    String name = JOptionPane.showInputDialog(null,"Welcome  Please enter your name:",titl,JOptionPane.PLAIN_MESSAGE,new ImageIcon(image0), null,"").toString();
    
    while(ages == true){
      int age = Integer.parseInt(JOptionPane.showInputDialog(null,"Welcome" + " " + name +"!" + " " + "How old are you?",titl2,3));     
    
      if(age <=28){
        JOptionPane.showMessageDialog(null,String.format("Text" + " " + "Your age:" + " " +age,titl,2));      
      }else if(age >=29 && age <=40){
        JOptionPane.showMessageDialog(null,String.format("Text" + " " + "Your age:"+ " "+age,titl,2));
      }else if(age>=41){
        JOptionPane.showMessageDialog(null,String.format("Text " + " " + "Your age:" + " "+age,titl,2));
      }else{
        // the part where I get stuck!
        // What to write here to catch an error and return the user to again input int age if he types String instead?
        continue;
      }
    }

I'm trying to validate the user input to allow only Int and if he types a String, he will get an error and again a window for age will pop-up.

I tried everything to catch the error, but I just keep failing at this. I could not accomplish this with age=Integer.parseInt(age) or .hasNextInt or other things like this. They always give me an error.

I saw many tutorials on how to do with normal Scanner and system.println, but I cannot figure out the resolution with JOptionPane.

I tried try/catch, but I don't get it and it never works.

Can you please help me? I'm new at Java and I would appreciate any help.

EDIT: I fixed it! Thank you Andrew Vershinin so much!

while(ages == true){
        Pattern AGE = Pattern.compile("[1-9][0-9]*");
    String  ageInput = JOptionPane.showInputDialog(null,"Welcome" + " " + name +"!" + " " + "How old are you?",titl2,3);
    if(!AGE.matcher(ageInput).matches()){
        JOptionPane.showMessageDialog(null,String.format("Please enter only numbers! :)"));
        continue;
    }
    int age = Integer.parseInt(ageInput);
    
    if(age <=28){
        JOptionPane.showMessageDialog(null,String.format("Text" + " " + "Your age:" + " " +age,titl,2));      
    }else if(age >=29 && age <=40){
        JOptionPane.showMessageDialog(null,String.format("Text" + " " + "Your age:"+ " "+age,titl,2));
    }else if(age>=41){
        JOptionPane.showMessageDialog(null,String.format("Text" + " " + "Your age:" + " "+age,titl,2));
    }

Andrew Vershinin · Accepted Answer · 2020-07-20 00:03:55Z

1

Using try/catch is a valid approach, but I would suggest using a regular expression, represented via a java.util.regex.Pattern instance: Pattern AGE = Pattern.compile("[1-9][0-9]*");, which could be translated as "a digit from 1 to 9, followed by any number of any digits", so it cannot be zero, and cannot contain leading zeroes.
Use it like this:

String ageInput = JOptionPane.showInputDialog(null,
    "Welcome" + " " + name +"!" + " " + "How old are you?",titl2,3)
if (!AGE.matcher(ageInput).matches()) {
    // handle invalid input
}

int age = Integer.parseInt(ageInput); // no exception there

answered Jul 20, 2020 at 0:03

Andrew Vershinin

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chunlong wang · Accepted Answer · 2020-07-20 04:02:02Z

0

Java built-in functions：

public static boolean isNumeric(String str){
for (int i = str.length();--i>=0;){ 
    if (!Character.isDigit(str.charAt(i))){
        return false;
    }
}
return true;

}

Regular expression, the fastest：

public static boolean isInteger(String str) { 
    Pattern pattern = Pattern.compile("^[-\\+]?[\\d]*$"); 
    return pattern.matcher(str).matches();

}

Use ascii code：

    public static boolean isNumeric(String str){
    for(int i=str.length();--i>=0;){
        int chr=str.charAt(i);
        if(chr<48 || chr>57)
            return false;
    }
   return true;
}

answered Jul 20, 2020 at 4:02

chunlong wang

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