我可以使用哪个类来读取Java中的整数变量?
答案 0 :(得分:133)
您可以使用java.util.Scanner
(API):
import java.util.Scanner;
//...
Scanner in = new Scanner(System.in);
int num = in.nextInt();
它还可以使用正则表达式等对输入进行标记.API有示例,此站点中还有许多其他示例(例如How do I keep a scanner from throwing exceptions when the wrong type is entered?)。
答案 1 :(得分:31)
如果您使用的是Java 6,则可以使用以下oneliner从控制台读取整数:
int n = Integer.parseInt(System.console().readLine());
答案 2 :(得分:17)
这里我提供2个例子来读取标准输入的整数值
示例1
import java.util.Scanner;
public class Maxof2
{
public static void main(String args[])
{
//taking value as command line argument.
Scanner in = new Scanner(System.in);
System.out.printf("Enter i Value: ");
int i = in.nextInt();
System.out.printf("Enter j Value: ");
int j = in.nextInt();
if(i > j)
System.out.println(i+"i is greater than "+j);
else
System.out.println(j+" is greater than "+i);
}
}
示例2
public class ReadandWritewhateveryoutype
{
public static void main(String args[]) throws java.lang.Exception
{
System.out.printf("This Program is used to Read and Write what ever you type \nType quit to Exit at any Moment\n\n");
java.io.BufferedReader r = new java.io.BufferedReader (new java.io.InputStreamReader (System.in));
String hi;
while (!(hi=r.readLine()).startsWith("quit"))System.out.printf("\nYou have typed: %s \n",hi);
}
}
我更喜欢第一个例子,它很容易理解 您可以在此网站上在线编译和运行JAVA程序:http://ideone.com
答案 3 :(得分:11)
检查一下:
public static void main(String[] args) {
String input = null;
int number = 0;
try {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
input = bufferedReader.readLine();
number = Integer.parseInt(input);
} catch (NumberFormatException ex) {
System.out.println("Not a number !");
} catch (IOException e) {
e.printStackTrace();
}
}
答案 4 :(得分:4)
上面的第二个答案是最简单的答案。
int n = Integer.parseInt(System.console().readLine());
问题是“如何从标准输入中读取”。
控制台是通常与启动程序的键盘和显示器相关联的设备。
您可能希望测试是否没有可用的Java控制台设备,例如Java VM无法从命令行启动,或者标准输入和输出流被重定向。
Console cons;
if ((cons = System.console()) == null) {
System.err.println("Unable to obtain console");
...
}
使用控制台是输入数字的简单方法。结合parseInt()/ Double()等。
s = cons.readLine("Enter a int: ");
int i = Integer.parseInt(s);
s = cons.readLine("Enter a double: ");
double d = Double.parseDouble(s);
答案 5 :(得分:4)
检查一下:
import java.io.*;
public class UserInputInteger
{
public static void main(String args[])throws IOException
{
InputStreamReader read = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(read);
int number;
System.out.println("Enter the number");
number = Integer.parseInt(in.readLine());
}
}
答案 6 :(得分:2)
这会引起麻烦,因此我更新了将使用2014年12月用户可用的最常用硬件和软件工具运行的解决方案。请注意JDK / SDK / JRE / Netbeans及其后续类,模板库编译器,编辑器和debuggerz是免费的。
该程序已使用Java v8 u25进行了测试。它是使用
编写和构建的
Netbeans IDE 8.0.2,JDK 1.8,操作系统是win8.1(道歉),浏览器是Chrome(双道歉)
- 旨在帮助UNIX-cmd-line OG处理基于GUI的基于Web的IDE
在ZERO COST - 因为信息(和IDE)应该始终是免费的。
通过Tapper7。对于每个人。
代码块:
package modchk; //Netbeans requirement.
import java.util.Scanner;
//import java.io.*; is not needed Netbeans automatically includes it.
public class Modchk {
public static void main(String[] args){
int input1;
int input2;
//Explicity define the purpose of the .exe to user:
System.out.println("Modchk by Tapper7. Tests IOStream and basic bool modulo fxn.\n"
+ "Commented and coded for C/C++ programmers new to Java\n");
//create an object that reads integers:
Scanner Cin = new Scanner(System.in);
//the following will throw() if you don't do you what it tells you or if
//int entered == ArrayIndex-out-of-bounds for your system. +-~2.1e9
System.out.println("Enter an integer wiseguy: ");
input1 = Cin.nextInt(); //this command emulates "cin >> input1;"
//I test like Ernie Banks played hardball: "Let's play two!"
System.out.println("Enter another integer...anyday now: ");
input2 = Cin.nextInt();
//debug the scanner and istream:
System.out.println("the 1st N entered by the user was " + input1);
System.out.println("the 2nd N entered by the user was " + input2);
//"do maths" on vars to make sure they are of use to me:
System.out.println("modchk for " + input1);
if(2 % input1 == 0){
System.out.print(input1 + " is even\n"); //<---same output effect as *.println
}else{
System.out.println(input1 + " is odd");
}//endif input1
//one mo' 'gain (as in istream dbg chk above)
System.out.println("modchk for " + input2);
if(2 % input2 == 0){
System.out.print(input2 + " is even\n");
}else{
System.out.println(input2 + " is odd");
}//endif input2
}//end main
}//end Modchk