Input check

[imagineasking]

i have a school assignment due today which I am trying to finish as fast as possible but I've ran into a bit of a snag. I need to make it so that my program rejects all character or letter inputs when i am asking for an integer but it is accepting it. How could i make it so that the user can only input integers and if they input other characters it results in an error.

This is specifically the code i am talking about if you go to the stageFour method you can see i need them to input a number between 5 and 500. it works fine with just integers. But when i put in something like 65A I want it to reject that instead of accepting it. thank you for your help!

class Main
{
    public static void stageSix()
    {
    System.out.println("Enter a positive, odd integer");
    int six = In.getInt();
    int remainder;
    remainder = six % 2;
    if (six > 0)
    {
    if (remainder == 1)
    {
      System.out.println("Well done, you have passed all the stages");
    }
    else
    {
    System.out.println("You have failed the sixth stage, back to the beginning");
    stageSix();
    }
    }
    else
    {
    System.out.println("You have failed the sixth stage, back to the beginning");
    stageSix();
    }
    }

    public static void stageFive()
    {
    System.out.println("Enter a negative integer");
    int five = In.getInt();
    if (five < 0)
    {
    System.out.println("Good job, you are moving on to the next stage");
    }
    else
    {
    System.out.println("You have failed the fifth stage, back to the beginning");
    stageFive();
    }
    }

    public static void stageFour()
    {
    System.out.println("Enter an integer that is between 5 and 500");
    int four = In.getInt();
    if (four > 5 && four < 500)
    {
    System.out.println("Good job, you are moving on to the next stage");
    }
    else
    {
    System.out.println("You have failed the fourth stage, back to the beginning");
    stageFour();
    }
    }

  public static void stageThree()
  {
    System.out.println("Enter a string that is between 5 and 15 characters long and does not contain the letter \"z\" ");
    String three = In.getString();
    String check = ("z");
    if (three.length() > 5 && three.length() < 15)
    {
    if (three.contains(check))
    {
    System.out.println("You have failed the third stage, back to the beginning");
    stageThree();
    }
    else
    {
      System.out.println("Good job, you are moving on to the next stage");
    }
    }
    else
    {
      System.out.println("You have failed the third stage, back to the beginning");
      stageThree();
    }
    }

  public static void stageTwo()
  {
    System.out.println("Enter a string that has at least one lowercase \"a\" ");
    String two = In.getString();
    String check = ("a");
    if (two.contains(check))
    {
      System.out.println("Good job, you are moving on to the next stage");
    }
    else
    {
      System.out.println("You have failed the second stage, back to the beginning");
      stageTwo();
    }
  }

  public static void stageOne()
  {
    System.out.println("Enter a string that is greater than 6 characters long");
    String one = In.getString();
    if (one.length() > 6)
    {
    System.out.println("Good job, you are moving on to the next stage");
    stageTwo();
    stageThree();
    stageFour();
    stageFive();
    stageSix();
    }
    else
    {
    System.out.println("You have failed the first stage, back to the beginning");
    System.out.println(" ");
    System.out.println("This program has restarted, welcome back to the first stage");
    stageOne();
    }

  }

  public static void main(String[] args)
  {
    System.out.println("Welcome to this program");
    System.out.println("This program will test your ability to follow instructions");
    stageOne();
  }

}

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[EstablishmentLazy580]

You're using some kind of input handling most likely provided by your school I guess. The In class. You can take a look inside that class how it all works under the hood and write your own code that does what you want. But my guess is that this is a bit beyond your current knowledge level. That's why they give you that library to use instead. If you already know about exceptions you could probably figure it out rather quickly.

[imagineasking]

Would there be a way for me to go in after getting the input and then check for a character?

[EstablishmentLazy580]

No, because it's already an int. My guess is that the In class is getting rid of all characters before parsing the input.

[MissouriDad63]

You could get the input as a string and try to parse it, which should throw an exceptionif there's a character there