Develop Reference

How do I go back to the menu after taking several questions?

I am currently creating a program where the user enters a specific set of questions. And the program must go back to the menu after completely answering all questions. How should I do it?
package com.company;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("""
\n \nAre you ready to take the quiz?
Enter "Y" to proceed or "N" to exit the program:""");
String TakeQuiz = input.nextLine();
if (TakeQuiz.equalsIgnoreCase("Y"))
do {
//blocks of code
}
}
}
System.out.println("Do you want to take the quiz again?");
String RetakeQuiz = input.nextLine();
while (RetakeQuiz.equalsIgnoreCase("Y")) ;
else {
System.out.println("We hope to see you again soon!");
System.exit(0);
}
}
}

There are many ways to achieve what you want, I would not clutter the main method and break the code to another function and loop there.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
for(;;)
takeQuiz();
}
public static void takeQuiz(){
Scanner input = new Scanner(System.in);
System.out.print("\n \nAre you ready to take the quiz?" +
"Enter \"Y\" to proceed or \"N\" to exit the program:");
String takeQuiz = input.nextLine();
if (takeQuiz.equalsIgnoreCase("Y")) {
System.out.println("Running code...");
System.out.println("Question 1");
System.out.println("Question 2");
System.out.println("Question 3");
}
// retake
if (takeQuiz.equalsIgnoreCase("R")){
takeQuiz();
}
if (takeQuiz.equalsIgnoreCase("N")){
System.out.println("We hope to see you again soon!");
System.exit(0);
}
}
}
Notice the escape character for quotes \" and the + for multiline Strings
Java 15 and beyond allows triple quotes as Java Text Blocks
so your String message should be valid

The basic structure is something like this:
boolean continueWithQuiz = true;
while (continueWithQuiz) {
// Put the code here for handling the quiz
...
// Should we keep going?
System.out.println("Do you want to take the quiz again?");
String retakeQuiz = input.nextLine();
continueWithQuiz = retakeQuiz == "Y";
}
One more comment. Please follow Java naming standards. Class names begin with an upper case letter. Constants should be ALL_CAPS. Everything else is in lower case.

My code isn't working

I am a beginner programmer and i am trying a program for my father.
import java.util.*;
import java.lang.*;
import java.io.*;
class Employee
{
String m1,m2,m3,m4,m5,m6,m7;
void main()
{
Scanner w=new Scanner(System.in);
Scanner n=new Scanner(System.in);
System.out.println("Please enter your name ");
String name=w.nextLine();
System.out.println("Please choose your client");
System.out.println("1 - XXXXXX");
int client=n.nextInt();
m1=name;//Storing name
if(client==1)//If statement storing client
{
m2="XXXXXX";
}
else
{
System.out.println("You have entered a wrong choice");
return;
}
String msg=m1+"\t"+m2;
System.out.println(msg);
}
}
This Code will give the output "as you have entered a wrong choice'"
It jumps to elsse statement. What is the error and is there an easier way to run this program. Thanks
Could yo please inform me on my error as

Ok try this code:
import java.util.Scanner;
public class Try
{
static String m1,m2,m3,m4,m5,m6,m7;
public static void main(String[] args)
{
Scanner w=new Scanner(System.in);
System.out.println("Please enter your name ");
String name=w.nextLine();
System.out.println("Please choose your client");
System.out.println("1 - XXXXXX");
int client=w.nextInt();
m1=name;//Storing name
if(client==1)//If statement storing client
{
m2="XXXXXX";
}
else
{
System.out.println("You have entered a wrong choice");
return;
}
String msg=m1+"\t"+m2;
System.out.println(msg);
}
}

You have missed you main method signature. In Java there is a specification of main method. Your main method should be like
public static void main(String []args){
}
In your case you main method should be
public static void main(String args[]) {
String m1, m2, m3, m4, m5, m6, m7;
Scanner w = new Scanner(System.in);
Scanner n = new Scanner(System.in);
System.out.println("Please enter your name ");
String name = w.nextLine();
System.out.println("Please choose your client");
System.out.println("1 - XXXXXX");
int client = n.nextInt();
m1 = name;//Storing name
if (client == 1)//If statement storing client
{
m2 = "XXXXXX";
} else {
System.out.println("You have entered a wrong choice");
return;
}
String msg = m1 + "\t" + m2;
System.out.println(msg);
}

Your problem are the 2 scanners.
Because a scanner work with an iterator, that keep the position inside the given inputstream (in this case), when you instantiate the 2 scanners, they both set their iterator at the same position into the stream, then you use "w.nextLine();", and the first scanner advances trough the stream returning the first line, as you wish, but the second scanner, that you haven't used, is still at the beginning of the stream, so basically when you use n.nextInt();, the scanner tries to parse your name as int, and it's strange that it doesn't throws an InputMismatchException, as it should do ("https://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#nextInt%28%29").
Rework your code as #Sarthak Mittal suggested and it should work.
PS: keep in mind indentation, it's important, really

First:
void main()
There is no such thing in Java. It should be,
public static void main(String[] args)
To know the meanings of public, static, String[] args read this: Explanation of 'String args[]' and static in 'public static void main(String[] args)'
Secondly,
int client = n.nextInt();
The value inside client depends on your input. If you input 2 or 3 instead of 1, your code'll definitely go to the else part. So make sure your input is 1.
Thirdly,
Get rid of the extra scanner. You need only one.
The rest of your code is ok.

how to avoid pressing several time enter affects the input data?

I'm writting a JAVA Class to validate input data, especifically integer numbers.
The class I develop is running fine but when I press more than one time enter and then a char type, it display several times " Error!! Invalid number. Try again. " and I would like to avoid it.
I have use nextLine() method but it doesn't seems to correct it.
Here is the Class:
package chapter07.libro;
import java.util.Scanner;
public class Validator_integer
{
public static int getInt (Scanner scanner, String promt)
{
int numberInteger = 0;
boolean isValid = false;
System.out.println(promt);
while(isValid == false)
{
if(scanner.hasNextInt())
{
numberInteger= scanner.nextInt();
isValid = true;
}//if
else
{
System.out.println("Error!! Invalid number. Try again.");
}//else
scanner.nextLine();
}//while
return numberInteger;
}//getInt
}//Validator_integer
and next is the app to use the class:
package chapter.prueba;
import java.util.Scanner;
import chapter07.libro.Validator_integer;
public class Test_Validator_Integer
{
public static void main(String[] args) {
Scanner sc = new Scanner (System.in);
String choice = "y";
while(choice.equalsIgnoreCase("y"))
{
int number = Validator_integer.getInt(sc, "Enter integer number: ");
System.out.println(number);
System.out.println("Continue (y/n): ");
choice = sc.next();
}//while
}//main
}//Test_Validator_Integer
The results I get are next:
Enter integer number:
2
2
Continue (y/n):
y
Enter integer number:
(Here I press several time enter)
xx
Error!! Invalid number. Try again.
Error!! Invalid number. Try again.
Error!! Invalid number. Try again.
Error!! Invalid number. Try again.
2
2
Continue (y/n):
n
So the part of (Error!! Invalid number. Try again.) displayed several times, is the one I would like to avoid.
Does any one know how to fix it???
Thanks in advance!!!

Before you read from System.in, make sure to "clear" it's contents to get rid of buffered/queued-up input. If there are n characters queued-up, then skip that many chars and you'll need to enter something new.
int n = System.in.available();
System.in.skip(n);
http://docs.oracle.com/javase/7/docs/api/java/io/InputStream.html

Reading an integer from the command prompt in Java using Scanner

I can't figure out what I'm doing wrong here.
My code:
import java.util.Scanner;
public class test
{
public static void main(String args[])
{
Scanner input_scanner = new Scanner(System.in);
System.out.printf("\nEnter number:");
int num_shapes = input_scanner.nextInt();
System.out.printf("\n%d", num_shapes);
}
}
Each time the program is run, I enter an integer, press the enter key, am taken to a new line, enter a new integer, and hit the enter key again. The first integer is then displayed.
How can I get it to display the first integer directly after it is entered, without having to enter a second integer?
I've tried it with and without
input_scanner.nextLine();
following the line containing 'nextInt()', but get the same thing either way.
Any help in resolving this problem is greatly appreciated.

this works perfectly
import java.util.Scanner;
public class mainclass{
public static void main(String args[])
{
Scanner input_scanner = new Scanner(System.in);
System.out.println("Enter number:");
int num_shapes = input_scanner.nextInt();
System.out.println(num_shapes);
}
}

Looping in Java(Beginners)

I have to use a loop in my code so that when someone enters yes, they can re-enter their names as many times as they want, but I have no idea how to do this. Any help is appreciated, here is my code:
public static void main(String [] args)
{
// Create a Scanner object to read input.
Scanner keyboard = new Scanner(System.in);
//Get the user's name.
System.out.print("What is your name?");
String name = keyboard.nextLine();
System.out.println("Hello there," + name);
System.out.println("Would you like to enter another name? Please enter Yes Or No.");
String reply = keyboard.nextLine();
if (reply == "yes")
{
}
}
}

This reply == "yes" is not how you compare Strings in Java. This compares there memory locations, not there contents (and it's unlikely there memory locations are going to be equal).
Instead you need to use reply.equals("yes") or if you don't care about doing a case comparison, you can use reply.equalsIgnoreCase("yes") instead
do {
// The remainder of your code...
} while (reply.equalsIgnoreCase("yes"));
Updated
You may also wish to have a read through The while and do-while statements and The for Statement, which covers the basics of looping in Java

Use a do-while loop:
public static void main(String[] args) {
// Create a Scanner object to read input.
Scanner keyboard = new Scanner(System.in);
do {
//Get the user's name.
System.out.print("What is your name?");
String name = keyboard.nextLine();
System.out.println("Hello there," + name);
System.out.println("Would you like to enter another name? Please enter Yes Or No.");
} while (keyboard.nextLine().equalsIgnoreCase("yes"));
System.out.println("Bye!");
keyboard.close();
}
}

use this
import java.util.*;
public class prob13 {
public static void main(String [] args)
{
// Create a Scanner object to read input.
Scanner keyboard = new Scanner(System.in);
//Get the user's name.
while(true){
System.out.print("What is your name?");
String name = keyboard.nextLine();
System.out.println("Hello there," + name);
System.out.println("Would you like to enter another name? Please enter Yes Or No.");
String reply = keyboard.nextLine();
if(reply.equals("no"))
break;
}
}
}
The reason for this is to loop through as long as the answer is not no.
or you could use this if you want the answer to always be yes
import java.util.*;
public class prob13 {
public static void main(String [] args)
{
// Create a Scanner object to read input.
Scanner keyboard = new Scanner(System.in);
String reply="yes";
//Get the user's name.
while(reply.equals("yes")){
System.out.print("What is your name?");
String name = keyboard.nextLine();
System.out.println("Hello there," + name);
System.out.println("Would you like to enter another name? Please enter Yes Or No.");
reply = keyboard.nextLine();
}
}
}

I think this will work (untested):
public static Scanner keyboard = new Scanner(System.in); // global
public static void main(String [] args)
{
getName();
}
public static void getName()
{
System.out.print("What is your name?");
String name = keyboard.nextLine();
System.out.println("Hello there," + name);
rerun();
}
public static void rerun()
{
System.out.println("Would you like to enter another name? Please enter \"yes\" or \"no\".");
String reply = keyboard.nextLine();
if (reply.equals("yes")) getName();
else System.exit();
}
}
First we call the getName() method and run through that once. Then we make a call to the rerun() method. This method will test if we want to re-run the program. If the user types in "yes", then we repeat the whole process. If we type in anything besides "yes", the program quits.
Besides the fact that your code is unfinished, the only real problem with your code is that you try to compare strings with the == operator. See MadProgrammer's answer as to why that is wrong.

The simplest (and probably clearest) way is to wrap what you want to repeat in a do-while statement:
public static void main(String [] args)
{
// Create a Scanner object to read input.
Scanner keyboard = new Scanner(System.in);
String reply;
do {
//Get the user's name.
System.out.print("What is your name?");
String name = keyboard.nextLine();
System.out.println("Hello there," + name);
System.out.println("Would you like to enter another name? Please enter Yes Or No.");
reply = keyboard.nextLine();
} while ("yes".equals(reply));
}
}
The reply variable must be declared before the block, because it is accessed in the loop condition (a variable is only visible in the block it is declared in, so if reply were declared in the loop, it would not be available to the loop condition).
I changed the loop condition because the == operator compares Strings by reference, i.e. it will check whether both sides point to the same String object. The equals method, in contrast, checks that the content of the Strings is equal (i.e. they contain the same characters in the same order).