无法将UserId解析为类型错误Java

Question

I am new to methods such as the public void stuff on Java.

If someone enters First Name as: John and surname: Smith. My program should display the UserID as jsmith. That all works fine when I put it all in the main method. I want it all on the same class (UserID) I also get an error message saying Insert assignment operator expression where it says:

process.userID;

Here's the code:

import java.util.Scanner;
public class UserID {

Scanner myScan = new Scanner (System.in);

char firstLetter;
String firstName;
String minCharacters;
String surname; 
String userID;

public void firstName () {
    System.out.print("Please enter your firstname: ");
    firstName = myScan.nextLine();
}

public void surname () {
    System.out.print("Please enter your surname: ");
    surname = myScan.nextLine();
}

public void userID (String userID) {
    firstLetter = firstName.charAt(0);
    minCharacters=surname.substring(0, 5);
    userID = firstLetter+minCharacters;
    System.out.print("User ID = " + userID.toLowerCase());
    System.out.println ("\nPlease keep a note of your User ID");

    myScan.close();
}

public static void main(String[] args) {
    String userID;
    userID process = new UserID();
    process.userID;

}

}

Answer 1

除了丹尼尔·斯坦利（Daniel Stanley）的答案之外，您没有在主方法中或在从命令行获取输入的userId（）方法中调用firstname（）和surname（）方法。

Answer 2

This line:

userID process = new UserID();

should be:

UserID process = new UserID(); //capital 'U'

and then:

process.userID("SomeString"); //add brackets

Answer 3

public void userID () {
firstLetter = firstName.charAt(0);

minCharacters=surname.length()>=5?surname.substring(0, 5):surname;
userID = firstLetter+minCharacters;
System.out.print("User ID = " + userID.toLowerCase());
System.out.println ("\nPlease keep a note of your User ID");

myScan.close();

}

public static void main(String[] args) {
UserID process = new UserID();
process.firstName();
process.surname();
process.userID();

}

Updated your code. It works. First issue was with UserID as Daniel pointed. Later, the surname can be less than 5 digit so it takes care of that. since you are using firstname and surname then you do not need to pass userId in the method.

Answer 4

Try this, Code have some logical and Java naming convention errors;

I have edited you code respectively.

public class UserID {
    Scanner myScan = new Scanner(System.in);
    char firstLetter;
    String firstName;
    String minCharacters;
    String surName;
    String userID;

    public void setFirstName() {
        System.out.print("Please enter your firstname: ");
        this.firstName = myScan.nextLine();
    }

    public void setSurName() {
        System.out.print("Please enter your surname: ");
        this.surName = myScan.nextLine();
    }

    public void generateUserID() {
        firstLetter = firstName.charAt(0);
        minCharacters = surName;
        userID = firstLetter + minCharacters;
        System.out.print("User ID = " + userID.toLowerCase());
        System.out.println("\nPlease keep a note of your User ID");

        myScan.close();
    }

    public static void main(String[] args) {
        String userID;
        UserID process = new UserID();
        process.setFirstName();
        process.setSurName();
        process.generateUserID();
    }
}

Feel free to ask anything.