我试图创建一个计算器,但我不能让它工作,因为我不知道如何获得用户输入。

如何在Java中获得用户输入?


当前回答

import java.util.Scanner;

public class userinput {
    public static void main(String[] args) {        
        Scanner input = new Scanner(System.in);

        System.out.print("Name : ");
        String name = input.next();
        System.out.print("Last Name : ");
        String lname = input.next();
        System.out.print("Age : ");
        byte age = input.nextByte();

        System.out.println(" " );
        System.out.println(" " );

        System.out.println("Firt Name: " + name);
        System.out.println("Last Name: " + lname);
        System.out.println("      Age: " + age);
    }
}

其他回答

import java.util.Scanner; 

class Daytwo{
    public static void main(String[] args){
        System.out.println("HelloWorld");

        Scanner reader = new Scanner(System.in);
        System.out.println("Enter the number ");

        int n = reader.nextInt();
        System.out.println("You entered " + n);

    }
}

您可以使用BufferedReader获取用户输入。

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String accStr;  

System.out.println("Enter your Account number: ");
accStr = br.readLine();

它将在accStr中存储一个String值,因此您必须使用Integer.parseInt将其解析为int。

int accInt = Integer.parseInt(accStr);
class ex1 {    
    public static void main(String args[]){
        int a, b, c;
        a = Integer.parseInt(args[0]);
        b = Integer.parseInt(args[1]);
        c = a + b;
        System.out.println("c = " + c);
    }
}
// Output  
javac ex1.java
java ex1 10 20 
c = 30
Scanner input=new Scanner(System.in);
int integer=input.nextInt();
String string=input.next();
long longInteger=input.nextLong();
Scanner input = new Scanner(System.in);
String inputval = input.next();