如何在 Java 中检查用户输入是 String、double 还是 long

How to check if user input is String, double or long in Java

提问by user4799681

I'm a beginner in java. I want to check first if the user input is String or Double or int. If it's String, double or a minus number, the user should be prompted to enter a valid int number　again. Only when the user entered a valid number should then the program jump to try. I've been thinking for hours and I come up with nothing useful.Please help, thank you!

我是java初学者。我想首先检查用户输入是 String 还是 Double 或 int。如果是 String、double 或负数，则应提示用户再次输入有效的 int 数。只有当用户输入一个有效数字时，程序才会跳转尝试。我想了几个小时，我想不出任何有用的东西。请帮忙，谢谢！

import java.util.InputMismatchException;
import java.util.Scanner;

public class Fizz {

public static void main(String[] args) {

    System.out.println("Please enter a number");

    Scanner scan = new Scanner(System.in);

    try {

        Integer i = scan.nextInt();

        if (i % 3 == 0 && (i % 5 == 0)) {
            System.out.println("FizzBuzz");
        } else if (i % 3 == 0) {
            System.out.println("Fizz");
        } else if (i % 5 == 0) {
            System.out.println("Buzz");
        } else {
            System.out.println(i + "は3と5の倍数ではありません。");
        }
    } catch (InputMismatchException e) {
        System.out.println("");

    } finally {
        scan.close();
    }

}

回答by TheLostMind

One simple fix is to read the entire line / user input as a String. Something like this should work. (Untested code) :

一个简单的解决方法是将整行/用户输入作为字符串读取。像这样的事情应该有效。（未经测试的代码）：

   String s=null;
   boolean validInput=false; 
   do{
      s= scannerInstance.nextLine();
      if(s.matches("\d+")){// checks if input only contains digits
       validInput=true;
      }
      else{
       // invalid input
     }
    }while(!validInput);

回答by Abhishek Chauhan

You can also use Integer.parseInt and then check that integer for non negativity. You can catch NumberFormatException if the input is string or a double.

您还可以使用 Integer.parseInt 然后检查该整数是否为非负值。如果输入是字符串或双精度数，您可以捕获 NumberFormatException。

Scanner scan = new Scanner(System.in);
try {
     String s = scan.nextLine();
     int x = Integer.parseInt(s);
}
catch(NumberFormatException ex)
{
}

回答by Discern

Try this one. I used some conditions to indicate the input.

试试这个。我使用了一些条件来指示输入。

Scanner scan = new Scanner(System.in);
String input = scan.nextLine();
int charCount = input.length();
boolean flag = false;
for(int x=0; x<charCount; x++){   
   for(int y=0; y<10; y++){
       if(input.charAt(x)==Integer.toString(y))
          flag = true;
       else{
          flag = false;
          break;
       }
   }
}
if(flag){
   if(scan.hasNextDouble())
      System.out.println("Input is Double");
   else
      System.out.println("Input is Integer");
}   
else
   System.out.println("Invalid Input. Please Input a number");

回答by Ian2thedv

Try this. It will prompt for input until an intgreater than 0 is entered:

尝试这个。它将提示输入，直到输入int大于 0为止：

System.out.println("Please enter a number");

try (Scanner scan = new Scanner(System.in)) {
    while (scan.hasNext()) {
        int number;
        if (scan.hasNextInt()) {
            number = scan.nextInt();
        } else {
            System.out.println("Please enter a valid number");
            scan.next();
            continue;
        }

        if (number < 0) {
            System.out.println("Please enter a number > 0");
            continue;           
        }

        //At this stage, the number is an int >= 0
        System.out.println("User entered: " + number);
        break;
    }
}

回答by Daman Singh

boolean valid = false;
double n = 0;
String userInput = "";
Scanner input = new Scanner(System.in);

while(!valid){
    System.out.println("Enter the number: ");
    userInput = input.nextLine();

    try{
        n = Double.parseDouble(userInput);
        valid = true;
    }
    catch (NumberFormatException ex){
        System.out.println("Enter the valid number.");
    }
}