您可以使用Scanner类或Console类

Console console = System.console();
String input = console.readLine("Enter input:");

你可以像这样使用BufferedReader获取用户输入:

    InputStreamReader inp = new InputStreamReader(System.in);
    BufferedReader br = new BufferedReader(inp);
    // you will need to import these things.

这就是应用它们的方法

    String name = br.readline(); 

因此，当用户在控制台中输入他的名字时，“字符串名称”将存储该信息。

如果它是一个你想要存储的数字，代码将像这样:

    int x = Integer.parseInt(br.readLine());

希望这能有所帮助!

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。

然后，Add在main()旁边抛出IOException

DataInputStream input = new DataInputStream(System.in);
System.out.print("Enter your name");
String name = input.readLine();

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);
    }
}

使用JOptionPane就可以实现。

Int a =JOptionPane.showInputDialog(null,"Enter number:");