java接受用户输入代码

inputstreamreader
2024-12-03 12:41:43
推荐回答(3个)
回答1:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/*
 * System.in 标准输入流。是从键盘获取数据的
 * 
 * 键盘录入数据:
 *  A:main方法的args接收参数。
 *  java HelloWorld hello world java
 *  B:Scanner(JDK5以后的)
 *  Scanner sc = new Scanner(System.in);
 *  String s = sc.nextLine();
 *  int x = sc.nextInt()
 *  C:通过字符缓冲流包装标准输入流实现
 *  BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
 */
public class SystemInDemo {
public static void main(String[] args) throws IOException {
// //获取标准输入流
// InputStream is = System.in;
// //我要一次获取一行行不行呢?
// //行。
// //怎么实现呢?
// //要想实现,首先你得知道一次读取一行数据的方法是哪个呢?
// //readLine()
// //而这个方法在哪个类中呢?
// //BufferedReader
// //所以,你这次应该创建BufferedReader的对象,但是底层还是的使用标准输入流
// // BufferedReader br = new BufferedReader(is);
// //按照我们的推想,现在应该可以了,但是却报错了
// //原因是:字符缓冲流只能针对字符流操作,而你现在是字节流,所以不能是用?
// //那么,我还就想使用了,请大家给我一个解决方案?
// //把字节流转换为字符流,然后在通过字符缓冲流操作
// InputStreamReader isr = new InputStreamReader(is);
// BufferedReader br= new BufferedReader(isr);
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

System.out.println("请输入一个字符串:");
String line = br.readLine();
System.out.println("你输入的字符串是:" + line);

System.out.println("请输入一个整数:");
// int i = Integer.parseInt(br.readLine());
line = br.readLine();
int i = Integer.parseInt(line);
System.out.println("你输入的整数是:" + i);
}
}

回答2:

这是让用户从控制台输入的例子: java.util.Scanner sc = new java.util.Scanner(System.in); String s = sc.nextLine();//这个就是用户输入的数据 int i = Integer.parseInt(s);//转换成整数 double d = Double.parseDouble(s);//转换成双精度实数

回答3:

package com.it;

import java.util.Scanner;

public class Test {

public static void main(String[] args) {

Scanner input = new Scanner(System.in);
System.out.println("输入字符串:");
String str = input.next();

System.out.println("输入int类型数字:");
int num = input.nextInt();

System.out.println("输入double类型数字:");
double num_ = input.nextDouble();

}

}

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