Question

我的任务是编写一个程序来检查输入字符串是否是有效数字。一个有效的数字是任何数字，有或没有+或 - 符号和一个小数，如：（+ 1 -1 1.0 1.0000000000 -1.0 -1.000000000）但是，当我输入像：（1.2.3）这样的东西时它起作用。我做错了什么

import java.util.Scanner;

public class parseNum {

  public static void main(String[] args) {
    System.out.println("Please enter a number:");
    Scanner input = new Scanner(System.in);
    String inStr = input.nextLine();
    int i;
    String state = "start";

    for (i=0; i<inStr.length(); i++) {
      if (state.equals("start")) {
        if (inStr.charAt(i) == '+' || inStr.charAt(i) == '-') {
          state = "afterSign";
          continue;
        }
      }
      // i assume this is where i messed up
      if(inStr.charAt(i) == ('.')){
        if(inStr.charAt(i+1) >= '0' && inStr.charAt(i+1) <= '9'){
        state = "accept";
        continue;
        }
        else{
        state = "reject";
        break;
        }
      }

      if (inStr.charAt(i) >= '0' && inStr.charAt(i) <= '9') {
        state = "accept";
        continue;
      }
      else {
        state = "reject";
        break;
      }
    }

    if (state.equals("accept"))
      System.out.println("Thank you");
    else
      System.out.println("Invalid input");

    input.close();
  }

}

Answer 1

在你的for循环之上声明一个int来计算点数，

int dotCount = 0;

接下来，每次找到一个点时计数

if(inStr.charAt(i) == ('.')){
            dotCount++; //THIS LINE
            if(inStr.charAt(i+1) >= '0' && inStr.charAt(i+1) <= '9'){
                state = "accept";
                continue;
            }
            else{
                state = "reject";
                break;
            }
        }

最后，添加if语句，如果它找到多于1，则拒绝。

   if(dotCount > 1) {
            state = "reject";
            break;
        }

Answer 2

见评论：

import java.util.Scanner;

public class ParseNum {//use java naming conventions 

    public static void main(String[] args) {

        System.out.println("Please enter a number:");
        Scanner input = new Scanner(System.in);
        String inStr = input.nextLine();
        int i =0;
        String state = "start";
        int dotCount = 0;

        //check for +/- outside the loop. It should only be 
        //at the beginning. This will cause number with multiple 
        //+/- signs to fail 
        if (inStr.charAt(0) == '+' || inStr.charAt(0) == '-') {
            state = "afterSign";
            i=1; 
        }

        for (i=0; i<inStr.length(); i++) {

            if(inStr.charAt(i) == ('.')){
                if(dotCount == 0){
                    state = "accept";
                    dotCount++;
                    continue;
                }
                else{//multiple dots fond 
                    state = "reject"; 
                    break;
                }
            }

            if (inStr.charAt(i) >= '0' && inStr.charAt(i) <= '9') {
                state = "accept";
                continue;
            }
            else {
                state = "reject";
                break;
            }
        }

        if (state.equals("accept"))
            System.out.println("Thank you");
        else
            System.out.println("Invalid input");

        input.close();
    }
}

请注意，具有多个0的数字（如+000009.8）被视为有效。
更简单和更好的实施方式是：

    public static void main(String[] args) {

        System.out.println("Please enter a number:");
        Scanner input = new Scanner(System.in);
        String inStr = input.nextLine();
        String state = "start";

        try {

            Double d = Double.valueOf(inStr);
            state = "accept";
        } catch (NumberFormatException ex) {

            state = "reject";
        }

        if (state.equals("accept")) {
            System.out.println("Thank you");
        } else {
            System.out.println("Invalid input");
        }

        input.close();
    }

如何让我允许的输入更加严格

2 个答案: