CODEWITHSUNDEEP
javavalidationcounter
user3537993
user3537993

Reputation: 29

Java Input Validation for Number Range and Numeric values only with counter

I am trying to complete a task and I am unsure what route to take. I have tried while, if, and a combination of statements and cannot get the input validation I need.

The code below is my method I am trying to setup to accomplish this.

private static int getNumericInput(String quantity) {
    int count = 0;
    String input;
    input = JOptionPane.showInputDialog(quantity);
    int range = Integer.parseInt(input);

    while ((range > 9 || range < 1) && (count < 2)) {
        JOptionPane.showMessageDialog(null, 
                "Sorry that input is not valid, please choose a quantity from 1-9");
        input = JOptionPane.showInputDialog(quantity);
        count++;
    }
    if (count == 2) {
        JOptionPane.showMessageDialog(null, 
                "Sorry you failed to input a valid response, terminating.");
        System.exit(0);
    }
    return range;
}

Upvotes: 2

Views: 18383

Answers (4)

Praz
Praz

Reputation: 1

As of JDK 1.8, you can use the temporal package to solve this, along-with the try-catch as shown by rest of the answers from this thread:

import java.time.temporal.ValueRange;
...
public static void main (String[] args){
ValueRange range = ValueRange.of(1,9);

    int num, counter = 0;

    do {
    System.out.print("Enter num: ");
    num = Integer.parseInt(scanner.next());
    if (range.isValidIntValue(num))
        System.out.println("InRange");
    else 
        System.out.println("OutOfRange");

    counter ++;
    } while (!range.isValidIntValue(num) && counter < 3);
...
}

Upvotes: 0

ug_
ug_

Reputation: 11440

As others have said to see if a String is a valid integer you catch a NumberFormatException.

try {
    int number = Integer.parseInt(input);
    // no exception thrown, that means its a valid Integer
} catch(NumberFormatException e) {
    // invalid Integer
}

However I would also like to point out a code change, this is a prefect example of a do while loop. Do while loops are great when you want to use a loop but run the condition at the end of the first iteration.

In your case you always want to take the user input. By evaluating the while loops condition after the first loop you can reduce some of that duplicate code you have to do prior to the loop. Consider the following code change. 

int count = 0;
String input;
int range;
do {
    input = JOptionPane.showInputDialog(quantity);
    try {
        range = Integer.parseInt(input);
    } catch(NumberFormatException e) {
        JOptionPane.showMessageDialog(null, "Sorry that input is not valid, please choose a quantity from 1-9");
        count++;
        // set the range outside the range so we go through the loop again.
        range = -1;
    }
} while((range > 9 || range < 1) && (count < 2));

if (count == 2) {
    JOptionPane.showMessageDialog(null, 
            "Sorry you failed to input a valid response, terminating.");
    System.exit(0);
}
return range;

Upvotes: 1

Asif Bhutto
Asif Bhutto

Reputation: 3994

If input string does not contain a valid number format in string then this line will throw NumberFormatException 

int range = Integer.parseInt(input)

You need to put it try catch block

try {
            Integer.parseInt("test");
        } catch (java.lang.NumberFormatException e) {
                count++; //allow user for next attempt.
              //    showInputDialog HERE
              if(count==3) {
                // show your msg here in JDIalog.
              System.exit(0);
           }
}

For 3 chances to input the correct information, you need to use loop, inside loop call your showInputDialog method

Upvotes: 0

Zoyd
Zoyd

Reputation: 3559

This line:

int range = Integer.parseInt(input);

appears before your while loop. So, you know how to convert the input to an int. The next step is to realize that you should do it each time the user gives you an input. You are almost there.

Upvotes: 0

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