我喜欢以下几点:

public String readLine(String tPromptString) {
    byte[] tBuffer = new byte[256];
    int tPos = 0;
    System.out.print(tPromptString);

    while(true) {
        byte tNextByte = readByte();
        if(tNextByte == 10) {
            return new String(tBuffer, 0, tPos);
        }

        if(tNextByte != 13) {
            tBuffer[tPos] = tNextByte;
            ++tPos;
        }
    }
}

例如，我会这样做:

String name = this.readLine("What is your name?")

在这里，程序要求用户输入一个数字。在此之后，程序打印数字的数字和数字的和。

import java.util.Scanner;

public class PrintNumber {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int num = 0;
        int sum = 0;

        System.out.println(
            "Please enter a number to show its digits");
        num = scan.nextInt();

        System.out.println(
            "Here are the digits and the sum of the digits");
        while (num > 0) {
            System.out.println("==>" + num % 10);
            sum += num % 10;
            num = num / 10;   
        }
        System.out.println("Sum is " + sum);            
    }
}

以下是一个更完善的公认答案，解决了两个常见需求:

重复收集用户输入，直到输入退出值 处理无效输入值(本例中是非整数)

Code

package inputTest;

import java.util.Scanner;
import java.util.InputMismatchException;

public class InputTest {
    public static void main(String args[]) {
        Scanner reader = new Scanner(System.in);
        System.out.println("Please enter integers. Type 0 to exit.");

        boolean done = false;
        while (!done) {
            System.out.print("Enter an integer: ");
            try {
                int n = reader.nextInt();
                if (n == 0) {
                    done = true;
                }
                else {
                    // do something with the input
                    System.out.println("\tThe number entered was: " + n);
                }
            }
            catch (InputMismatchException e) {
                System.out.println("\tInvalid input type (must be an integer)");
                reader.nextLine();  // Clear invalid input from scanner buffer.
            }
        }
        System.out.println("Exiting...");
        reader.close();
    }
}

例子

Please enter integers. Type 0 to exit.
Enter an integer: 12
    The number entered was: 12
Enter an integer: -56
    The number entered was: -56
Enter an integer: 4.2
    Invalid input type (must be an integer)
Enter an integer: but i hate integers
    Invalid input type (must be an integer)
Enter an integer: 3
    The number entered was: 3
Enter an integer: 0
Exiting...

注意，如果没有nextLine()，错误的输入将在无限循环中重复触发相同的异常。您可能希望根据具体情况使用next()，但要知道像这样带有空格的输入将生成多个异常。

最简单的方法之一是使用Scanner对象，如下所示:

import java.util.Scanner;

Scanner reader = new Scanner(System.in);  // Reading from System.in
System.out.println("Enter a number: ");
int n = reader.nextInt(); // Scans the next token of the input as an int.
//once finished
reader.close();

你可以做一个简单的程序，要求用户的名字和打印任何回复使用输入。

或者让用户输入两个数字，你可以加、乘、减或除这些数字，并打印出用户输入的答案，就像计算器的行为一样。

你需要Scanner类。你必须导入java。util。scanner;在你需要使用的代码中

Scanner input = new Scanner(System.in);

Input是一个变量名。

Scanner input = new Scanner(System.in);

System.out.println("Please enter your name : ");
s = input.next(); // getting a String value

System.out.println("Please enter your age : ");
i = input.nextInt(); // getting an integer

System.out.println("Please enter your salary : ");
d = input.nextDouble(); // getting a double

看看这有什么不同:input.next();， i = input.nextInt();， d = input.nextDouble();

根据String类型，int和double类型以同样的方式变化。不要忘记代码顶部的import语句。

请参见博客文章“扫描器类和获取用户输入”。

在java中输入是非常简单的，你所要做的就是:

import java.util.Scanner;

class GetInputFromUser
{
    public static void main(String args[])
    {
        int a;
        float b;
        String s;

        Scanner in = new Scanner(System.in);

        System.out.println("Enter a string");
        s = in.nextLine();
        System.out.println("You entered string " + s);

        System.out.println("Enter an integer");
        a = in.nextInt();
        System.out.println("You entered integer " + a);

        System.out.println("Enter a float");
        b = in.nextFloat();
        System.out.println("You entered float " + b);
    }
}