如何验证扫描仪输入的 int 长度？

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希望它是一个 3 位数的整数。我无法将它放到正确的位置：

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);

这就是我需要实现它的地方，我遇到了这个问题。我敢肯定，这可能只是我需要的一条线不合适或缺少，我只是不知道是什么或在哪里。当它坐着时，它会不断地要求我输入一个 3 位数的数字，无论长度如何。对输入的字符串进行错误检查目前有效：

     //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);

剩下的就是做我需要做的事情了：

     //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 3 位数字：
2345
请输入 3 位数字：
12
请输入 3 位数字：
123
请输入 3 位数字：
motu
这不是一个三位数的数字！再试一次！

Answer 1

在 MvG 和 pingul 的帮助下，这就是目前我希望的工作方式：

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]);

   }
 }
}

Answer 2

压缩并重新格式化您的代码读取

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);

与the Scanner documentation 相比，我们可以看到scan.next() 调用将读取（并丢弃）非整数令牌。否则 j 将是您读取的整数。当你读到的数字与 3 不同时，你继续这样做。不是数字的长度，而是数字本身。所以如果你想结束循环，输入3。如果您想按照提示执行此操作，请输入003。

如果这不是您要检查的内容，请更改循环结束条件。或者也许改变您测试三位数字的方式，通过使用正则表达式来匹配这些数字。

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 cmets 正确指出的那样，模式 "\\d\\d\\d" 也可以写成 "[0-9]{3}"、"\\d{3}" 或 "[0-9][0-9][0-9]"。在位数是变量的情况下，使用 {…} 可能很有用。

documentation for Scanner.hasNext(Pattern) 需要模式来匹配输入。这显然遵循将整个字符串与模式匹配的Matcher.matches() 语义，而不是检查字符串是否包含与模式匹配的任何部分的Matcher.find()。所以输入不必像我最初假设的那样包含在^ 和$ 中，实际上不应该使用这些，除非模式是用Pattern.MULTILINE flag 编译的。

您可能想调用Scanner.useDelimiter 以仅使用换行符进行分隔。

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