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How to verify OTP in Java using simple loop?

I'm a Java beginner and my project consists of creating a simple program to register users for an alumni center. The process creates an ID and then provides the new user with an OTP. Next is the login (Enter ID:, Enter OTP: ).
My OTP verification method is not working. It seems to be a problem with the IF.equals declaration, the process jumps straight to the ELSE condition.
Any suggestions why?
Here is my code:
class Main {
static NewRegandLogin newRegAndLogin = new NewRegandLogin(null, null, null, null, null, null);
static ArrayList<NewRegandLogin> loginInformation = new ArrayList<>();
public static void main(String[] args) {
System.out.println(" WELCOME TO THE ALUMNI SHE-CODES SYSTEM ");
System.out.println("_________________________________\n - New Alumni registration - \n");
System.out.println("");
newRegAndLogin.registerNewGrad();
System.out.println();
System.out.println("_________________________________");
System.out.println();
System.out.println("Your new Alumni ID is: " + newRegAndLogin.getAlumniId());
System.out.println();
System.out.println("Your temporary password is:");
System.out.println(newRegAndLogin.oTp(8));
loginInformation.add(newRegAndLogin);
System.out.println("_________________________________");
System.out.println("_________________________________\n - Alumni Login - \n");
System.out.println("");
newRegAndLogin.login();
System.out.println("");
System.out.println("Please make a list of completed Courses: -->Enter 'S' to stop adding courses<--");
newRegAndLogin.setAlumniCourses();
System.out.println("_________________________________");
newRegAndLogin.setLinkedInPage();
loginInformation.add(newRegAndLogin);
//printAlumniProfile();
System.out.println("_________________________________");
newRegAndLogin.jobOffer();
}
void login() {
System.out.print("ID: ");
alumniIdImput = scanner.nextLine();
idVerification();
do {
System.out.println("Password (OTP if logging in for the first time): ");
passwordImput = scanner.nextLine();
oTpFromImput = passwordImput.toCharArray();
oTpVerification();
} while (isPasswordCorrect=false);
void oTpVerification() {
isPasswordCorrect = false;
if (oTpFromImput.equals(oTp(8))) {
isPasswordCorrect = true;
System.out.println("Logging In.....");
}else {
isPasswordCorrect = false;
System.out.println("Incorrect password.\nPlease enter valid password: 8 alpha numeric
characters(Aa,123,#,#,$,%)");
}
}
This is the oTp method
char[] oTp (int length) {
String capitalChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
String smallChars = "abcdefghijklmnopqrstuvwxyz";
String numbers = "0123456789";
String symbols = "!##$%^&*_-=+/.?<>";
String values = capitalChars + smallChars + numbers + symbols;
Random oneTimePassword = new Random();
char[] password = new char[length];
for(int i = 0; i<length;i++) {
password[i] = values.charAt(oneTimePassword.nextInt(values.length()));
}
return password;
}

It seems you built a guessing game, not an OTP verification code.
You first read the OTP from user, and only then randomly generate one to copare to it.
Basically, you code expects the user to guess a random 8 character password that has not been created you, which is basically impossible...
You need to generate to OTP first, show it to the user, then ask them to input it.

I see your logic code is generate OTP code after User input. It seem so wierd bro.
Whenever you call oTp(8) function will generate new OTP.
Use should generate OTP first then store somewhere, then User input and compare it.

You need to store the generated otp somewhere. Then compare it with the input otp. Right now you are comparing it with the otp(8). And otp(8) always returns a new otp.

Splitting a Scanner Input Into Strings

I've been looking for an answer to this for a while, but for some reason, none of them seem to work.
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Please enter full name (last, first)");
String[] personalInfo = scanner.next().split(", ");
String firstName = personalInfo[1];
String lastName = personalInfo[0];
System.out.println("Your info: " + firstName + " " + lastName);
There is my code. I'm basically trying to obtain the personal info, which would be the first and last name. I want to split the first and last name into 2 different strings, but whenever I try to print this, I get the error:
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: Index 1 > out of bounds for length 1
at Fines.main(Fines.java:11)
I'm confused because I even started the array with 0 like I was supposed to.. I just don't understand what is going incorrectly.
Please give me a hand - thanks in advance!

What you want is scanner.nextLine() to read from standard input up until end of line. Then split would work as you expected.
Scanner scanner = new Scanner(System.in);
System.out.println("Please enter full name (last, first)");
String[] personalInfo = scanner.nextLine().split(", ");
String firstName = personalInfo[1];
String lastName = personalInfo[0];
System.out.println("Your info: " + firstName + " " + lastName);

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException:
Index 1 > out of bounds for length 1 at Fines.main(Fines.java:11)
As the size of the personalInfo is 1 not 2.
use nextLine() instead of next() because next() will only return the input that comes before a space.
String[] personalInfo = scanner.next().split(", "); should be
String[] personalInfo = scanner.nextLine().split(", ");
You might want to read this What's the difference between next() and nextLine() methods from Scanner class?

try (Scanner scan = new Scanner(System.in)) {
System.out.println("Please enter full name (last, first)");
String firstName = scan.next();
String lastName = scan.next();
System.out.println("Your info: " + firstName + ' ' + lastName);
}
scanner.next() read until next delimiter (space by default), so it ready only the firstName. Just replace it with scanner.nextLine() or use scanner.next() two times.

Read spaces in String as Input Without using NextLine

Discription:
I used several scanner statments in below code. First i Read the "Adress" and then "Mobile No." and then some random stuff from user.
When i use adress=sc.next()
it reads the string value of adress from user(Without space) and go to the Next scan statement i.e. mobile=sc.nextBigInteger(). by using this method i am not able read the spaces in "adress(string) " and it throws the runtime error as inputMismatchException.
Now if i use the adress=sc.NextLine, the Progrram directly jumps to the mobile=sc.nextBigInteger().
How is it possible to read the spaces as input in above situations and the below code. How can i protect myself from runtime Errors. I got similar questions on the forum but non of them was satisfactory. Thank you. (Ask me if you need more information about question)
Expected:
input(In adress string) : pune maharashtra india
output(in display function) : pune mahrashtra india.
What exactly happened: if input(in adress string) : pune output(in display function) : pune
if input(in adress string) : Pune india
(Now As soon as i entered the string and hit the enter there I get a runtime error as an inputmismatchexception )
> JAVA
public class Data {
Scanner sc = new Scanner(System.in);
String adress;
BigInteger AcNo;
BigInteger mobile;
String ifsc;
void getData() {
System.out.println("Welcome to Bank System");
System.out.println("Please Enter the adress :");
adress= sc.next();
System.out.println("Enter the Mobile Number");
mobile = sc.nextBigInteger();
System.out.println("Enter the Account Number");
AcNo = sc.nextBigInteger();
System.out.println("Enter the IFSC Code");
ifsc= sc.next();
}
public static void main(String[] args) {
Data d=new Data();
d.getData();
}
}

Change this line ;
adress= sc.next();
with this line ;
adress = sc.nextLine();
this will solve your problem. scan.nextLine() returns everything up until the next new line delimiter, or \n ,but scan.next() is not.
So all code will be like this ;
public class Data{
String adress;
BigInteger AcNo;
BigInteger mobile;
String ifsc;
void getData() {
System.out.println("Welcome to Bank System");
System.out.println("Please Enter the adress :");
adress = new Scanner(System.in).nextLine();
System.out.println("Enter the Mobile Number");
mobile = new Scanner(System.in).nextBigInteger();
System.out.println("Enter the Account Number");
AcNo = new Scanner(System.in).nextBigInteger();
System.out.println("Enter the IFSC Code");
ifsc = new Scanner(System.in).next();
}
public static void main(String[] args) {
Data d = new Data();
d.getData();
}
}

Java GetstringMethod

The assignment:
Write a program (Greetings) that prompts the user to enter the first name, the last name, and year of birth, then it returns a greetings message
in proper format (see the example below).
Create a method(s) that accept the scanner and a prompt as parameters and return the user input. A separate method should accept the user input results as parameters, format and print the results. No print statement or scanner input should happen inside main(). Here is an example dialogue with the user:
Please enter your first name:
tom
Please enter your last name:
cruise
Please enter your year of birth:
1962
Greetings, T. Cruise! You are about 53 years old.
I finished the code, but right now it is giving me a compilation error. How do i fix it?
import java.util.*;
public class Greetings {
public static void main(String[] args) {
Scanner newscanner = new Scanner(System.in);
String ask = ("Please enter your first name: ");
String ask2 = ("Please enter your last name: ");
String ask3 = ("Please enter your year of birth: ");
public static String getString(Scanner newscanner, String ask, String ask2, String ask3){
System.out.println(ask);
String first = newscanner.next();
String firstletter = first.substring(0,1).toUpperCase() ;
return firstletter;
System.out.println(ask2);
String second = newscanner.next();
int x = second.length();
String y = second.substring(0, x).toLowerCase();
String lastname = y.substring(0,1).toUpperCase();
return lastname;
System.out.println(ask3);
int third = newscanner.nextInt();
int age = (2015 - third);
return age
System.out.println("Greetings, "+ firstletter + ". " + lastname+"!" +" You are about " + age + " years old");
}
}
}

Hard to read, but I think you actually have the getString() method inside your main() method - it needs to be after it, and only be called from inside main(), not defined there.

I need help on the Java Scanner Code

So I need help on this code. This code is all in one so ignore the spaces but I need to write another scanner in the way bottom of the code and if I do add
String feeling = in.nextLine(); at the very end it does not work. I need a it so that I can write my feelings so that I can make jarvis answer but the string does not work and java ignores the string and goes right on to the next part. It starts from the middle.
Scanner in = new Scanner(System.in);
System.out.println("Type User Name:");
String userName = in.nextLine();
System.out.println("PASSWORD:");
int passcodeFromUser=in.nextInt();
int passcode = 2015;
if (passcodeFromUser == passcode) {
System.out.println("Welcome Mr." + userName + "!");
Random random = new Random(userName.hashCode());
System.out.println("Mr." + userName + ", You are now recognized and you are now able to command me.");
System.out.println("I was created by John Choi");
System.out.println("JARVIS stands for Just A Rather Very Intelligent System");
System.out.println("How are you today Mr." + userName + "?");
}
So if I add this code at the back it does not work. It ignores and says Oh. Mr is feeling.
String feeling = in.nextLine();
System.out.println("Oh. Mr." + userName + "is feeling" + feeling + ".")

That is because your nextInt invocation does not actually parse a line feed.
Quoting the API, Scanner#nextInt:
Scans the next token of the input as an int.
(focus on the token part here)
Here's one (but not the only) way to fix it:
Integer passcodeFromUser = null;
try {
passcodeFromUser= Integer.parseInt(in.nextLine());
}
catch (NumberFormatException nfe) {
// TODO handle non-numeric password
}
... instead of int passcodeFromUser=in.nextInt();.
You can also loop the parsing of the Integer so that you print an error message when catching the NumberFormatException and don't break the loop until you have a valid numeric passcode.

You can consume the \n character:
in.nextLine();
String feeling = in.nextLine();
So just putting in.nextLine() before the code you were going to add will easily fix your problem.