Java 向量循环输入错误直到正确

Question

我需要输入一个用户输入的数量，只能在1到60之间的数字，小于1和大于60的值应该是无效的...并要求重新输入。

问题是，我不知道如何重复问题并将插入错误值的矢量方块替换为正确值。

例如：它一直在询问值，忽略它的错误并稍后显示... 例如vector[6]，最后的值显示为：[45, 34, 23, 22, 11, 99] or [45, 55, 34, 99, 99, 99]

我的任务：开发一种记录赌注的算法。玩家可以执行下注 6 到 15 个数字，介于 1 到 60 之间。然后算法必须要求通知用户：

您下注的号码数量；
接收投注（1到60之间）；
按顺序显示下注；
显示要为此收取的金额（在 table 我得到之后）。

我的尝试：

Scanner teclado = new Scanner(System.in);
System.out.println("How many bets from 6 to 15?");
int qntd = teclado.nextInt();
int c;
int i = 0;
if ((qntd < 6) || (qntd > 15)){
    System.out.println("Only 6 to 15 values.");
} else {
    int[] bets = new int [qntd]; 
    for (c=0; c<=bets.length-1; c++ ){
        i++;
        System.out.println("Choose a number ");
        bets[c] = teclado.nextInt();               
        if ((bets[c] < 1) || (bets[c]>60)) {
            System.out.println("Only 1 to 60 values.");
        }    
    }    
    Arrays.sort(bets);
    System.out.print("Realized bets [");
    for (int valor: bets){
        System.out.print( " " + valor  );                         
    }
    System.out.println(" ]");
    System.out.println(" ]");
    String pr = null;   
    int vlr = qntd;                     
    switch (vlr){
        case 6 : 
            pr = "4,50.";
            break;
        case 7 : 
            pr = "31,50.";
            break;
        case 8 : 
            pr = "126,00.";
            break;
        case 9 : 
            pr = "378,00.";
            break;
         case 10 : 
            pr = "945,00.";
            break;
        case 11 : 
            pr = "2.079,00.";
            break;
        case 12 : 
            pr = "4.158,00.";
            break;
        case 13:
            pr = "7.722,00,";
            break;
        case 14 :
            pr = "13.513,50.";
            break;
        case 15 : 
            pr = "22.522,50.";                                 
    }
    System.out.println("Bets value is R$ " + pr);
}

Answer 1

您不必向数组中插入无效输入。

int c=0;
while(c<bets.length){
    System.out.println("Choose a number ");
    int input = teclado.nextInt();
    if ((input < 1) || (input > 60)) {
        System.out.println("Only 1 to 60 values.");
        continue;
    }
    bets[c]=input;  
    c++; 
}

对于无效输入，循环将继续而不将输入添加到数组。此外 c 只会针对有效输入递增。

Answer 2

Scanner teclado = new Scanner(System.in);
  System.out.println("How many bets from 6 to 15?");
  int qntd = 0;
  int c;
  int i = 0;
  while(!(qntd >6 && qntd < 15))
  {
    qntd = teclado.nextInt();
    if(!(qntd >=6 && qntd <= 15))
    {
      System.out.println("Only 6 to 15 values.");
    }
  }
  int[] bets = new int [qntd];
  for (c=0; c<=qntd-1; c++ )
  {
    bets[c] = 0;
    while((bets[c] < 1) || (bets[c]>60))
    {
      bets[c] = teclado.nextInt();
      if ((bets[c] < 1) || (bets[c]>60)) 
      {
        System.out.println("Only 1 to 60 values.");
      }   
    }
  }
    Arrays.sort(bets);
                    System.out.print("Realized bets [");
               for (int valor: bets){
                 System.out.print( " " + valor  );
                 
               }
               System.out.println(" ]");
               System.out.println(" ]");
            String pr = null;   
            int vlr = qntd;   
            
            switch (vlr){
        case 6 : 
            pr = "4,50.";
           break;
        case 7 : 
            pr = "31,50.";
            break;
        case 8 : 
            pr = "126,00.";
            break;
        case 9 : 
            pr = "378,00.";
            break;
        case 10 : 
            pr = "945,00.";
            break;
        case 11 : 
            pr = "2.079,00.";
            break;
        case 12 : 
            pr = "4.158,00.";
            break;
        case 13:
            pr = "7.722,00,";
            break;
        case 14 :
            pr = "13.513,50.";
            break;
        case 15 : 
            pr = "22.522,50.";
           
        
    }
    System.out.println("Bets value is R$ " + pr);
  }

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