我怎样才能使这个输出更容易？

Question

我遇到的问题是我必须输入50个整数，而当我点击空格时它将不会识别该号码，并且当我点击输入时，那么我的输出框就会非常长，只有1个数字＃的

static final int MAX = 50; 
public static void main(String[] args) 
{ 
    int index; 
    int check; 
    int infants = 0, children = 0, teens = 0, adults = 0; 
    System.out.println("This program will count the number of people " 
         + "and how many of that age group cameto the college fair."); 
    System.out.println("**********************************************************"); 
    System.out.println("Please enter the integer value data:"); 
    int [] peopleTypes = new int[MAX]; 

    Scanner keyboard = new Scanner(System.in); 

    for(index=0; index<MAX; index++) 
     { 
      peopleTypes[index] = keyboard.nextInt(); 
      if(peopleTypes[index] == 1) 
       infants = infants + 1; 
      if(peopleTypes[index] == 2) 
       children = children + 1; 
      if(peopleTypes[index] == 3) 
       teens = teens + 1; 
      if(peopleTypes[index] == 4) 
       adults = adults + 1; 
      else 
       index = index-1; 
      System.out.print(""); 
     } 

    System.out.println("The number of infants that were at the college fair was: " + infants); 
    System.out.println("The number of children that were at the college fair was: " + children); 
    System.out.println("The number of teenagers that were at the college fair was: " + teens); 
    System.out.println("The number of adults that were at the college fair was: " + adults); 

Answer 1

你可以读你输入的字符串，并尝试分析它：

String s = ""; 
int i = 0; 
for (index = 0; index < MAX; index++) { 
    s = keyboard.next(); 
    try { 
     i = Integer.valueOf(s.trim()); 
    } catch (NumberFormatException nfe) { 
     // log exception if needed 
     System.out.println(s + " is not a valid integer."); 
     continue; 
    } 

    // do the rest 
    peopleTypes[index] = i; 

    //... 
} 

Answer 2

尝试使用此：

public class ScannerDelimiter { 
    public static void main(String[] args) { 
     Scanner scanner = new Scanner("12, 42, 78, 99, 42"); 
     scanner.useDelimiter("\\s*,\\s*"); 
     while (scanner.hasNextInt()) { 
      System.out.println(scanner.nextInt()); 
     } 
    } 
} /* Output: 
12 
42 
78 
99 
42 

在这种情况下，分隔符是

<any number of spaces or tabs>,<any number of spaces or tabs>

Answer 3

Java中的控制台输入是行缓冲的。直到用户输入，你才会得到任何东西。打空间，只是增加一个空间，你会得到的线。注意：点击退格删除一个你不会看到的字符。