Problems and solutions encountered in using nextline(), nextint() and next() in scanner

One 、 problem

Code ：

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

public class JavaDemo {
    
    public static void main(String[] args) {
    
        // Read... From the keyboard 
        Scanner sc = new Scanner(System.in);
        // establish ArrayList A collection of objects 
        ArrayList<Student> students = new ArrayList<>();

        // Loop input 5 Student information , And deposit in ArrayList Collection 
        for (int i = 0; i < 5; i++) {
    
            System.out.println(" Please enter the first " + (i + 1) + " Names of students ：");
            String name = sc.nextLine();
            System.out.println(" Please enter the first " + (i + 1) + " Student gender ：");
            String gender = sc.nextLine();
            System.out.println(" Please enter the first " + (i + 1) + " Age of students ：");
            int age = sc.nextInt();
            Student student = new Student(name, gender, age);
            students.add(student);
        }

        // Print 
        for (Student s : students) {
    
            System.out.println(s);
        }
    }
}

The operation results are as follows ：

As you can see from the results , Entering the second student name was skipped .

The reason is that nextLine() and nextInt() The ending way of is different ,nextLine() The method is to enter \n end ,nextInt() The method is Hit enter \n Or the space ends , Only read the input before entering int Type value , And continue to pass the following bytes .

This leads to nextLine() When the method receives the first loop nextInt() Enter after input \n, That is to say “23\n” Medium “\n”, Although enter \n No display , But there is still transmission , So enter the number 2 When there are student names, the program will read enter \n end , Make the next entry .

Two 、 use next() solve

To solve this problem , Just put nextLine() Switch to next() Just go , because next() End with a space , Cannot read carriage return \n Or space , And encounter the next carriage return \n Or the space will stop .

Code ：

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

public class JavaDemo {
    
    public static void main(String[] args) {
    
        // Read... From the keyboard 
        Scanner sc = new Scanner(System.in);
        // establish ArrayList A collection of objects 
        ArrayList<Student> students = new ArrayList<>();

        // Loop input 5 Student information , And deposit in ArrayList Collection 
        for (int i = 0; i < 5; i++) {
    
            System.out.println(" Please enter the first " + (i + 1) + " Names of students ：");
            String name = sc.next();
            System.out.println(" Please enter the first " + (i + 1) + " Student gender ：");
            String gender = sc.next();
            System.out.println(" Please enter the first " + (i + 1) + " Age of students ：");
            int age = sc.nextInt();
            Student student = new Student(name, gender, age);
            students.add(student);
        }

        // Print 
        for (Student s : students) {
    
            System.out.println(s);
        }
    }
}

The operation results are as follows ：