Question

我能够让程序运行并使用错误检查来确保用户输入实际上是一个int。我遇到的问题是我只想要它是一个3位数的int。我很难把它带到正确的地方：

import java.util.*;

public class listMnemonics

{

   public static void main(String[] args)

   {

   //Defines the "keypad" similar to that of a phone
   char[][] letters = 

   {{'0'},{'1'},{'A','B','C'},{'D','E','F'},{'G','H','I'},{'J','K','L'}, 
   {'M','N','O'},{'P','Q','R','S'},{'T','U','V'},{'W','X','Y','Z'}};

     //Creates the Scanner
     Scanner scan = new Scanner(System.in);

这就是我需要实现的地方，我遇到了这个问题。我确定它可能只是一条不合适的地方或我想要的遗失，我只是不知道在哪里或哪里。无论长度如何，它都会不断地让我输入一个3位数字。检查输入的字符串的错误当前有效：

     //Gives instructions to the user to enter 3-digit number
     //Any amount of numbers will work, but instructions help
     //System.out.println("Please enter a 3-digit number: ");

     int j;


     do
     {
     System.out.println("Please enter a 3-digit number: ");

           while (!scan.hasNextInt()) {

     System.out.println("That's not a 3-digit number! Try again!");
     scan.next(); // this is important!
     }

     j = scan.nextInt();

     }



     //while (j <= 0); This works while not checking digit length
     while (j != 3);
     int w = (int) Math.log10(j) +1; //Found this, but not sure if it helps or not

     String n = Integer.toString(w);

以下是其他可以做到我需要做的事情：

     //Determines char length based on user input
     char[][] sel = new char[n.length()][];

     for (int i = 0; i < n.length(); i++)

     {

        //Grabs the characters at their given position
        int digit = Integer.parseInt("" +n.charAt(i));
        sel[i] = letters[digit];

     }

  mnemonics(sel, 0, "");

}

public static void mnemonics(char[][] symbols, int n,  String s) 

{

  if (n == symbols.length)

  {

     System.out.println(s);
     return;

  }

  for (int i = 0; i < symbols[n].length; i ++) 

  {

     mnemonics(symbols, n+1, s + symbols[n][i]);

  }
 }
}

这是输出：

---- jGRASP exec：java listMnemonics
   请输入一个3位数字：
    2345
   请输入一个3位数字：
    12
   请输入一个3位数字：
    123
   请输入一个3位数字：
    MOTU
   这不是一个3位数的数字！再试一次！

Answer 1

压缩并重新格式化您的代码

Scanner scan = new Scanner(System.in);
int j;
do {
  System.out.println("Please enter a 3-digit number: ");
  while (!scan.hasNextInt()) {
     System.out.println("That's not a 3-digit number! Try again!");
     scan.next(); // this is important!
   }
   j = scan.nextInt();
} while (j != 3);

将其与the Scanner documentation进行比较，我们可以看到scan.next()调用将读取（并丢弃）非int令牌。否则j将是您读取的整数。并且你继续这样做，而你读的数字不同于3.不是数字的长度，而是数字本身。因此，如果您想结束循环，请输入3。如果您想在按照提示操作时这样做，请输入003。

如果这不是您要检查的内容，则更改循环结束条件。或者通过使用正则表达式匹配这些数字来改变测试三位数字的方式。

Scanner scan = new Scanner(System.in);
Pattern threeDigitNumber = Pattern.compile("\\d\\d\\d");
int j;
System.out.println("Please enter a 3-digit number: ");
while (!scan.hasNext(threeDigitNumber)) {
  if (scan.hasNext()) {
    System.out.println(scan.next() + " is not a 3-digit number! Try again!");
  } else {
    System.out.println("Input terminated unepxectedly");
    System.exit(1);
  }
}
j = scan.nextInt();

正如评论所指出的那样，模式"\\d\\d\\d"也可以写为"[0-9]{3}"，或"\\d{3}"或"[0-9][0-9][0-9]"。在数字位数是变量的情况下，使用{…}可能很有用。

documentation for Scanner.hasNext(Pattern)要求模式匹配输入。这显然遵循将Matcher.matches()语义与模式匹配整个字符串，而不是Matcher.find()来检查字符串是否包含与模式匹配的任何部分。因此，输入不必包含在^和$中，正如我最初假设的那样，实际上不应该使用这些输入，除非使用Pattern.MULTILINE flag编译模式。 / p>

您可能希望仅使用换行符调用Scanner.useDelimiter来划分界限。

Scanner.useDelimiter("[\\r\\n]+")

Answer 2

在MvG和pingul的帮助下，这正是我目前希望的工作方式：

import java.util.*;
import java.util.regex.Pattern;

public class listMnemonics

{

   public static void main(String[] args)

{

   // Defines the "keypad" similar to that of a phone
   char[][] letters = 

   {{'0'},{'1'},{'A','B','C'},{'D','E','F'},{'G','H','I'},{'J','K','L'}, 
   {'M','N','O'},{'P','Q','R','S'},{'T','U','V'},{'W','X','Y','Z'}};

      // Creates the Scanner
      Scanner scan = new Scanner(System.in);

      // Gives instructions to the user to enter 3-digit number
      // This 'Pattern' also guarantees that only 3 digits works.

      Pattern threeDigitNumber = Pattern.compile("[0-9]{3}");

      int j;

      do

      {

         System.out.println("Please enter a 3-digit phone number: ");

         // If it's not a 3-digit int, try again
         while (!scan.hasNext(threeDigitNumber)) 

            {

            System.out.println("That's not a 3-digit number! Try again!");

            // This is important!
            scan.next(); 

            }

         j = scan.nextInt();

      }

      while (j <= 0);

      String n = Integer.toString(j);


      // Determines char length based on user input
      char[][] sel = new char[n.length()][];

      for (int i = 0; i < n.length(); i++)

      {

         // Grabs the characters at their given position
         int digit = Integer.parseInt("" +n.charAt(i));
         sel[i] = letters[digit];

      }

   mnemonics(sel, 0, "");

}

// Here is where the magic happens and creates the possible
// letter combinations based on the user input and characters
// selected in previous steps.
public static void mnemonics(char[][] symbols, int n,  String s) 

{

   if (n == symbols.length)

   {

      System.out.println(s);
      return;

   }

   for (int i = 0; i < symbols[n].length; i ++) 

   {

      mnemonics(symbols, n+1, s + symbols[n][i]);

   }
 }
}

如何验证扫描仪输入的int长度？

2 个答案: