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Java 循环扫描器输入直到输入或“e”没有.split

[英]Java loop scanner input until enter or “e” without .split


public class test {

   private static Scanner userpress = new Scanner(System.in);

   public static void main(String[] args) {
      int r = 0;
      int h = 0;

      System.out.println("---------------------------------");
      System.out.println("write your two numbers (numerator, denominator)");

      userpress.useDelimiter("\\s"); // here
      while (userpress.hasNextInt()) {

         r = userpress.nextInt();
         h = userpress.nextInt();
         // userpress.nextLine(); // remove
         int x = r / h;
         System.out.print(x + " ");

      }

   }

   private static void message() {
      System.out.println("user pressed e");

   }

}

This program asks the user for minimum of 2 inputs, the first input is numerator and the second will be denominator, then it will give the value of numerator/denominator and print it out.该程序要求用户至少输入 2 个输入,第一个输入是分子,第二个输入是分母,然后它会给出分子/分母的值并打印出来。 Here is what i want to be able to do: I want to be able to write down as many numbers as possible, for example 10 5 20 4 , the output should be 2, 5 and if i write 10 5 20 4 30 5 the output should be 2, 5, 6 .这是我想要做的:我希望能够写下尽可能多的数字,例如10 5 20 4 ,output 应该是2, 5如果我写10 5 20 4 30 5 output 应该是2, 5, 6 however this only works for even numbers.但是这只适用于偶数。 if i write uneven numbers for example 10 5 3 , then the program crashes because 2 is not being divided by anything.如果我写奇数,例如10 5 3 ,那么程序会崩溃,因为 2 没有被任何东西除。 I want the program to delete the last input if its uneven, for example if input is 10 5 3 , then output should be 2 since 10/5=2 and the last input should be disgarded.如果最后一个输入不均匀,我希望程序删除它,例如如果输入是10 5 3 ,那么 output 应该是2 ,因为 10/5=2 并且最后一个输入应该被忽略。 How do I solve that issue?我该如何解决这个问题? problem number 2: If i write an e after the numbers, for example;问题2:例如,如果我在数字后写一个e 10 5 20 4 e , then I want the output to be 2, 5 user pressed e I dont want to use.split and i have to use hasnext... and next() 10 5 20 4 e ,然后我希望 output 为2, 5 user pressed e我不想使用.split,我必须使用hasnext...next()

You can use regex for checking spaces and than take the only first part before space.您可以使用正则表达式来检查空格,而不是在空格之前取唯一的第一部分。 Like this:像这样:

  import java.util.Scanner;

  public class Main {

    private static Scanner userpress = new Scanner(System.in);

    public static void main(String[] args) {
        String first;
        String second;

        System.out.println("---------------------------------");
        System.out.println("write your two numbers (numerator, denominator)");

        userpress.useDelimiter("\\s"); // here
        while (userpress.hasNextLine()) {

            first = userpress.nextLine();
            second = userpress.nextLine();
            if(first.contains(" ") || second.contains(" ")) {
                String[] firstParts = first.split(" ");
                String[] secondParts = second.split(" ");
                first = firstParts[0];
                second = secondParts[0];
            }
            // userpress.nextLine(); // remove
            int x = Integer.valueOf(first) / Integer.valueOf(second);
            System.out.print(x + " ");

        }

    }

    private static void message() {
        System.out.println("user pressed e");

    }

}

The problem with your code is that you are scanning two integers with a single check of userpress.hasNextInt() .您的代码的问题是您通过一次检查userpress.hasNextInt()来扫描两个整数。 This check should be there for every scan of an integer.每次扫描 integer 时都应进行此检查。

The working code:工作代码:

import java.util.Scanner;

public class Main {
    public static void main(String args[]) {
        int r = 0;
        int h = 0;
        System.out.print("Write your two numbers (numerator, denominator): ");
        Scanner userpress = new Scanner(System.in);
        userpress.useDelimiter("\\s");

        while (userpress.hasNextInt()) {
            r = userpress.nextInt();
            if (userpress.hasNextInt()) {
                h = userpress.nextInt();
                int x = r / h;
                System.out.print(x + " ");
            } else {
                break;
            }
        }
        String s = userpress.next();
        if ("e".equalsIgnoreCase(s)) {
            message();
        }
    }

    private static void message() {
        System.out.println("user pressed e");
    }
}

A sample run:示例运行:

Write your two numbers (numerator, denominator): 10 5 3 e
2 user pressed e

Another sample run:另一个示例运行:

Write your two numbers (numerator, denominator): 10 5 3
2 

Another sample run:另一个示例运行:

Write your two numbers (numerator, denominator): 10 5 20 5
2 4 

Another sample run:另一个示例运行:

Write your two numbers (numerator, denominator): 10 5 20 5 e
2 4 user pressed e

Alternatively, if you would not mind using a little bit more space, you could come up with a simple algorithm, like:或者,如果您不介意使用更多空间,您可以提出一个简单的算法,例如:

public class Main {

    private static Scanner userpress = new Scanner(System.in);

    public static void main(String[] args) {
        System.out.println("---------------------------------");
        System.out.println("write your two numbers (numerator, denominator)");
        userpress.useDelimiter("\\s");

        List<Integer> numbers = new ArrayList<>();

        while(userpress.hasNextInt()) {
            numbers.add(userpress.nextInt());
        }

        for (int i=0; i<numbers.size()-1; i+=2) {
            int x = numbers.get(i)/numbers.get(i+1);
            System.out.print(x + " ");
        }
    }
}

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