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
在这里,程序要求用户输入一个数字。在此之后,程序打印数字的数字和数字的和。
import java.util.Scanner;
public class PrintNumber {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int num = 0;
int sum = 0;
System.out.println(
"Please enter a number to show its digits");
num = scan.nextInt();
System.out.println(
"Here are the digits and the sum of the digits");
while (num > 0) {
System.out.println("==>" + num % 10);
sum += num % 10;
num = num / 10;
}
System.out.println("Sum is " + sum);
}
}
其他回答
Scanner input=new Scanner(System.in); int integer=input.nextInt(); String string=input.next(); long longInteger=input.nextLong();
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。
import java.util.Scanner;
public class userinput {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Name : ");
String name = input.next();
System.out.print("Last Name : ");
String lname = input.next();
System.out.print("Age : ");
byte age = input.nextByte();
System.out.println(" " );
System.out.println(" " );
System.out.println("Firt Name: " + name);
System.out.println("Last Name: " + lname);
System.out.println(" Age: " + age);
}
}
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();
您可以使用BufferedReader获取用户输入。
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String accStr;
System.out.println("Enter your Account number: ");
accStr = br.readLine();
它将在accStr中存储一个String值,因此您必须使用Integer.parseInt将其解析为int。
int accInt = Integer.parseInt(accStr);