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
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...");
}
}
}
}
其他回答
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。
您可以根据需求使用以下任何选项。
扫描仪类
import java.util.Scanner;
//...
Scanner scan = new Scanner(System.in);
String s = scan.next();
int i = scan.nextInt();
BufferedReader和InputStreamReader类
import java.io.BufferedReader;
import java.io.InputStreamReader;
//...
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
int i = Integer.parseInt(s);
DataInputStream类
import java.io.DataInputStream;
//...
DataInputStream dis = new DataInputStream(System.in);
int i = dis.readInt();
DataInputStream类中的readLine方法已弃用。要获得String值,您应该使用前面的BufferedReader解决方案
控制台类
import java.io.Console;
//...
Console console = System.console();
String s = console.readLine();
int i = Integer.parseInt(console.readLine());
显然,这种方法在某些ide中不能很好地工作。
下面是获取键盘输入的方法:
Scanner scanner = new Scanner (System.in);
System.out.print("Enter your name");
String name = scanner.next(); // Get what the user types.
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();
下面是使用java.util.Scanner的问题中的程序:
import java.util.Scanner;
public class Example {
public static void main(String[] args) {
int input = 0;
System.out.println("The super insano calculator");
System.out.println("enter the corrosponding number:");
Scanner reader3 = new Scanner(System.in);
System.out.println(
"1. Add | 2. Subtract | 3. Divide | 4. Multiply");
input = reader3.nextInt();
int a = 0, b = 0;
Scanner reader = new Scanner(System.in);
System.out.println("Enter the first number");
// get user input for a
a = reader.nextInt();
Scanner reader1 = new Scanner(System.in);
System.out.println("Enter the scend number");
// get user input for b
b = reader1.nextInt();
switch (input){
case 1: System.out.println(a + " + " + b + " = " + add(a, b));
break;
case 2: System.out.println(a + " - " + b + " = " + subtract(a, b));
break;
case 3: System.out.println(a + " / " + b + " = " + divide(a, b));
break;
case 4: System.out.println(a + " * " + b + " = " + multiply(a, b));
break;
default: System.out.println("your input is invalid!");
break;
}
}
static int add(int lhs, int rhs) { return lhs + rhs; }
static int subtract(int lhs, int rhs) { return lhs - rhs; }
static int divide(int lhs, int rhs) { return lhs / rhs; }
static int multiply(int lhs, int rhs) { return lhs * rhs; }
}