How to use do-while loop to check user input again?

Question

I want to add a loop to my program so that when a user enters an incorrect name it goes back to the start of the program and asks them to enter their name again. I think I need a do-while loop but I am not sure how to implement it with the if statements and boolean already included. I want the user to be only have three entries and if they get it wrong three times then the program closes.

import java.util.Scanner;

public class Username
{
  public static void main(String[] args)
  {
    {
      Scanner kb = new Scanner(System.in);
      // array containing usernames
      String[] name = {"barry", "matty", "olly", "joey"}; // elements in array


      System.out.println("Enter your name");
      String name1 = kb.nextLine();
      boolean b = true;
      for (int i = 0; i < name.length; i++)
      {
        if (name[i].equals(name1))
        {
          System.out.println("you are verified you may use the lift");
          b = false;
          break;// to stop loop checking names
        }
      }

      if (b)
      {
        System.out.println("Invalid entry 2 attempts remaining, try again");
      }
    }

man, few minutes ago, you asked a question. I gave you an answer and now, you pasted my code here and you are asking again... — ruhungry
– ruhungry, Commented Mar 29, 2014 at 16:29
@GirlyGirl i think you're right, we are doing his assignment :S. — Salah
– Salah, Commented Mar 29, 2014 at 16:35

ylun · Accepted Answer · 2014-03-29 16:27:00Z

1

You can use a condition in the while loop. Something along the lines of:

boolean b = false;
while(!b){
    System.out.println("Enter your name");
    String name1 = kb.nextLine();
    for (int i = 0; i < name.length; i++) {
        if (name[i].equals(name1)) {
            b = true;
            System.out.println("you are verified you may use the lift");
        }else{
            System.out.println("Invalid entry 2 attempts remaining, try again");
        }
    }
}

The loop will quit if the name condition is fulfilled and will loop around if it is not.

answered Mar 29, 2014 at 16:27

ylun

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Comments

Salah · Accepted Answer · 2014-03-29 16:27:19Z

1

You can do it like this:

int count = 0;
point:
do {
    System.out.println("Enter your name");
    String name1 = kb.nextLine();
    boolean b = true;
    for (int i = 0; i < name.length; i++) {
        if (name[i].equals(name1)) {

            System.out.println("you are verified you may use the lift");
            b = false;
            break point;// to stop loop checking names
        }
    }

    if (b) {
        count++;
        System.out.println("Invalid entry 2 attempts remaining, try again");
    }
while(!b || count <=3)

answered Mar 29, 2014 at 16:27

Salah

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Comments

Aman Agnihotri · Accepted Answer · 2014-03-29 16:50:45Z

Use the following approach. Good thing is that it is a clean and robust solution.

import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class AccessPoint
{
  private Scanner scanner;
  private List<String> usernames;

  public AccessPoint()
  {
    scanner = new Scanner(System.in);
    usernames = Arrays.asList("Barry", "Matty", "Olly", "Joey");

    if (tryAccessForTimes(3))
    {
      allowAccess();
    }
    else
    {
      denyAccess();
    }

    scanner.close();
  }

  public static void main(String[] args)
  {
    new AccessPoint();
  }

  private boolean tryAccessForTimes(int times)
  {
    boolean accessAllowed = false;

    for (int tryIndex = 1; tryIndex <= times && !accessAllowed; tryIndex++)
    {
      String userInput = getUserName();

      for (String userName : usernames)
      {
        if (userName.equals(userInput))
        {
          accessAllowed = true;
          break;
        }
      }

      if (!accessAllowed)
      {
        printNumberOfTriesLeft(times, tryIndex);
      }
    }

    return accessAllowed;
  }

  private void printNumberOfTriesLeft(int times, int tryIndex)
  {
    int triesLeft = times - tryIndex;

    if (triesLeft != 0)
    {
      System.out.println("You have " + triesLeft
        + (triesLeft == 1 ? " try" : " tries") + " left.");
    }
  }

  private String getUserName()
  {
    System.out.print("Enter Username: ");
    return scanner.nextLine();
  }

  private void allowAccess()
  {
    System.out.println("Access Granted. Allowed to use lift.");
  }

  private void denyAccess()
  {
    System.out.println("Access Denied.");
  }
}

loknath · Accepted Answer · 2014-03-29 17:33:32Z

1

package com.loknath.lab;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class User1 {
public static void main(String[] args) {

    Scanner kb = new Scanner(System.in);
    // array containing usernames
    String[] name = {"zerr", "barry", "matty", "olly", "joey" }; // elements
    String []temp=name;
    Arrays.sort(temp);
    while (true) {

        System.out.println("Enter your name");
        String name1 = kb.nextLine();

        if (Arrays.binarySearch(temp,name1)>=0) {
            System.out.println("you are verified you may use the lift");
            break;
        } else {
            System.out.println("Not a verified user try again!");
        }

    }
    System.out.println("Done");
}

 }

output

  Enter your name
  loknath
  Not a verified user try again!
  Enter your name
  chiku
  Not a verified user try again!
  Enter your name
  zerr
  you are verified you may use the lift
  Done

edited Mar 29, 2014 at 17:33

answered Mar 29, 2014 at 16:38

loknath

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7 Comments

Aman Agnihotri Over a year ago

The binarySearch will work only if the elements in the name array are sorted. The question says about no such constraint.

loknath Over a year ago

its return +ve value if the element present inside the array else it return the -ve

Aman Agnihotri Over a year ago

@loknath: Make your name array as : String[] name = { "olly", "matty", "joey", "barry"}; or any other unsorted name list for the matter, and your binarySearch based program will not run properly.

Aman Agnihotri Over a year ago

@loknath: Read my above comment again. Your binarySearch approach will fail if you make the name array as an unsorted array. Its incidentally giving correct answer because the name array is currently sorted, but may not be when the name array changes.

loknath Over a year ago

@AmanAgnihotri binarySeach method return index of the search key, if it is contained in the array; otherwise, (-(insertion point) - 1). so this method return either +ve in case the search key found else it ll return -ve

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