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?
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
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();
其他回答
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
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
System.out.println("Welcome to the best program in the world! ");
while (true) {
System.out.print("Enter a query: ");
Scanner scan = new Scanner(System.in);
String s = scan.nextLine();
if (s.equals("q")) {
System.out.println("The program is ending now ....");
break;
} else {
System.out.println("The program is running...");
}
}
}
}
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?")
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();
Scanner input = new Scanner(System.in);
String inputval = input.next();