Question

Scanner keyboard = new Scanner(System.in);
System.out.println("Enter number of bars(min 1 / max 6): ");
int barNumbers = keyboard.nextInt();
while (barNumbers <= 0 || barNumbers >= 7) {
  System.out.println("Enter number of bars(min 1 / max 6): ");
  barNumbers = keyboard.nextInt();
}
int barHeights[] = new int[barNumbers];
int countForBars = 1;
for (int i = 0; i <= barNumbers - 1; i++) {
  System.out.println(
      "Enter height of bar " + countForBars + " of " + barNumbers + "(min 1 / max 7): ");
  barHeights[i] = keyboard.nextInt();
  while (barHeights[i] <= 0 || barHeights[i] >= 8) {
    System.out.println(
        "Enter height of bar " + countForBars + " of " + barNumbers + "(min 1 / max 7): ");
    barHeights[i] = keyboard.nextInt();
  }
  countForBars += 1;
}

大家好，我需要一些帮助，这对你来说很容易，因为我是Java的初学者。无论如何，我现在试图控制一个机器人手臂通过Java传达东西。问题是，正如你在那里看到的那样，我设置了一些条件，只能获得1-6之间的整数。因此必须限制其余的输入，尤其是字符串。那么如何限制字符串输入而不是使用try catch东西呢？如果有任何有用的解决方案，我会很感激。干杯!抱歉代码指示。我不知道如何让它像代码一样:(

Answer 1

可能您可以尝试以下代码：

 public static boolean isInteger(String str) { //pass your "keyboard" string here.
 if (str == null) { // if there is nothing entered in the String
    return false;
 }
 int length = str.length();
 if (length == 0) {
    return false;
 }
 int i = 0;
 if (str.charAt(0) == '-') {
    if (length == 1) {
        return false;
    }

 }
for (i = 1; i < length; i++) {
    char c = str.charAt(i);
    if (c < '0' || c > '6') {
        return false;
     }
   }
return true;
}

如何在Java中不尝试catch时接受整数输入

1 个答案: