java我需要从输入中提取前N个字符

Question

我刚刚开始学习Java。我需要从输入中提取前N个字符。输入包括输入字符串开头的字符数，然后是字符数。输出包括开头的数字

例如，如果用户输入“3stars”，输出将是“3st”，或者输入“7Appendments”，输出将是“7append”

尝试了以下方法：

  Scanner input = new Scanner(System.in);
      System.out.print("Please enter a uncoded string: ");
      String first = input.nextLine();
      input.close();

      if (first.charAt(0) == 'u') {
          first = first.toUpperCase();
      } else if (first.charAt(0) == 'l') {
          first = first.toLowerCase();
      } else if (first.charAt(0) == 'e') {
            String str = "";
            for (int i = 0; i < first.length(); i = i + 2) {
                str += first.charAt(i);
            }
            first = str;
        } else if (first.charAt(0) == 'o') {
            String str = "";
            for (int i = 1; i < first.length(); i = i + 2) {
                str += first.charAt(i);
            }
            first = str;
        } else if (first.charAt(0) == 1++) {
            String str = "";
            for (int i = 1++ ; i < first.length(i); i = i charAt(i)) {
                str += first.charAt(i);
            }
            first = str;

任何帮助都会很好

Answer 1

我希望这有助于：

int num = Character.getNumericValue(first.charAt(0));

String str = "";
for(int i = 0;i < num;i++){
   str += first.charAt(i);
}
System.out.println(str);

Answer 2

这也适用于11abcdefghijklmnopqrstuvwxyz等数字大于9的输入

public static void main(String[] args) {
    String input = "7appendices";
    // match numbers
    Pattern p = Pattern.compile("^([0-9]+)");
    // by default at least the integer has one digit
    int digits = 1;
    int length = 0;
    // while the beginning of the input match an integer
    while (p.matcher(input.substring(0, digits)).matches()) {
        // set length as the integer at the beginning
        length = Integer.parseInt(input.substring(0, digits));
        // increase the amount of digits from the integer in the input
        digits++;
    }
    System.out.println(input.substring(0, length));
}

你只需要像这样用scanner替换输入

 String input = new Scanner(System.in).next();