Java代码未按计划运行

Question

This isn't the entire code, just the problem area. (Full code below)

if (passLength > 4) {
    System.out.println("Signup alost complete.");

    Random rand = new Random();

    int randm = rand.nextInt(100000) + 1;

    System.out.println(
            "To confirm you are not a robot, please enter this code: "
                    + randm);

    String code = userInput.next();

    if (code.equals(randm)) {
        System.out.println("Thank you, " + userName);
        System.out.println(
                "You may now login. Begin by entering your username");

        if (userInput.equals(userName)) {
            System.out.println("Now please enter you password");
        }

        // Where the rest of the program will go
    }

    else {
        System.out.println("The code entered is incorrect.");
    }

}

else {
    System.out.println("Invalid Password");
}

I am making a program where the user makes an account, then later signs in. The part I am having trouble with is a verification to ensure the user is human (they obviously are, but still). After creating their username and password, I generate and print a random int, and they have to type it back.

My problem is that the program always skips to the else statement, and prints "The code entered is incorrect." Even when I enter it perfectly.

Can anyone tell me what I'm doing wrong?

Below is the entire code, just in case.

public static void main(String[] args) {

    System.out.println("Hi! To begin please choose your username.");
    System.out.println("To do this, please enter your username below. ");
    System.out.println("This name must be at least three characters.");

    Scanner userInput = new Scanner(System.in);

    String userName = userInput.next();

    int nameLength = userName.length();

    if (nameLength > 2) {

        System.out.println("Now please enter your password.");
        System.out
                .println("This password must be at lease five characters");

        String passWord = userInput.next();

        int passLength = passWord.length();

        if (passLength > 4) {
            System.out.println("Signup alost complete.");

            Random rand = new Random();

            int randm = rand.nextInt(100000) + 1;

            System.out.println(
                    "To confirm you are not a robot, please enter this code: "
                            + randm);

            String code = userInput.next();

            if (code.equals(randm)) {
                System.out.println("Thank you, " + userName);
                System.out.println(
                        "You may now login. Begin by entering your username");

                if (userInput.equals(userName)) {
                    System.out.println("Now please enter you password");
                }

                // Where the rest of the program will go
            }

            else {
                System.out.println("The code entered is incorrect.");
            }

        }

        else {
            System.out.println("Invalid Password");
        }

    }

    else {
        System.out.println("Invalid Username");

    }

}

Answer 1

You are comparing a String with an integer, which can't be the same obviously.

int randm = rand.nextInt(100000) + 1;
...
String code = userInput.next();
if  (code.equals(randm)) 

To solve the problem you could convert the integer to a String and compare the strings (or the other way).

String randm = String.valueOfrand.nextInt(100000) + 1;
...
String code = userInput.next();
if  (code.equals(randm)) // Comparing string now

Edit : As @Pshemo pointed out, int code = userInput.nextInt(); will do the work given that you compare the integers with == .

Answer 2

It's because randm is int and code is String. So the code if (code.equals(randm)) always results to false.

Answer 3

You cannot compare a string and an integer. Trying taking the input as an integer instead of a String. String code = userInput.next();

Instead use:

int code= userInput.nextInt();
if(code==randm)

Or you could convert the integer to a String as well