Question

Scanner scanner = null;
try {
    Scanner myfile = new Scanner(new File("postfix.txt"));
    while (myfile.hasNext()) {
        input = myfile.nextLine();
        inputArray = input.split(" ");
        if (inputArray.length == 3) {
            inputArray = input.split(" ");
            Double number1 = Double.parseDouble(inputArray[0]);
            Double number2 = Double.parseDouble(inputArray[1]);
            String operator = inputArray[2];
            try {
                switch (operator) {
                case "+":
                    result = number1 + number2;
                    break;
                case "-":
                    result = number1 - number2;
                    break;
                case "/":
                    result = number1 / number2;
                    break;
                case "*":
                    result = number1 * number2;
                    break;
                default:
                    System.out.println("Received unsupported operator: "
                            + operator);
                    break;
                }
                System.out.println("Your expression is:" + number1
                        + operator + number2);
                System.out.println("Your answer is: " + result);
            } catch (NumberFormatException e) {
                System.err.println("Invalid expression");
                continue;
            }
        }
    }
} catch (FileNotFoundException e) {
    System.out.println("Your file is not found: -" + e.getMessage());
}

文字就像这样

44 3 +
9.99 + 0.09
12 0 *
. 10 -
10.2 2 *
12 4 /
66.1 0.12 -
.0 99.10 +
300 4.0 +
* 20 10 /
10 20
5.2 +
1 2 &
100 139 -
- 80 2
9 5 2
4 / 3
3 A -
200.5 10 *
2 * 4
8 2 *
10 20 -
8 16 /
-4 12 +
+ 4 2
x y z

表达式：

Please enter your file name:

所有上述表达式都应该解决，但是那些表达式如下： x y z, + 4 2应该提出无效的表达。

当我运行我的代码时，文本中只有一个表达式被解析。之后，它会出现无效的表达，但不会继续阅读文本。此外，它需要要求用户键入文件名并从中读取，但此刻它只从系统中已有的文件中读取。

Answer 1

主要问题是你问的文件的第二行双倍运算符，你有双运算符双

9.99 + 0.09

以防止使用try chatch

停止解析程序

设置

try{...

解析之前

用于从控制台读取文件名

Scanner lineScanner = new Scanner(System.in);
        System.out.println("Please enter your file name: ");
        String path = lineScanner.nextLine();
        Scanner myfile = new Scanner(new File(path));....

Answer 2

这里试试这个，

    Scanner scanner = null;
    try {
        Scanner myfile = new Scanner(new File("postfix.txt"));
        while (myfile.hasNext()) {
            String input = myfile.nextLine();
            inputArray = input.split(" ");
            if (inputArray.length == 3) {
                inputArray = input.split(" ");

                String operator = null;
                Double number1 = null;
                Double number2 = null;
                try {
                    number1 = Double.parseDouble(inputArray[0]);
                    number2 = Double.parseDouble(inputArray[1]);
                    operator = inputArray[2];
                } catch (NumberFormatException numberFormatException) {
                    continue;
                }
                try {
                    switch (operator) {
                        case "+":
                            result = number1 + number2;
                            break;
                        case "-":
                            result = number1 - number2;
                            break;
                        case "/":
                            result = number1 / number2;
                            break;
                        case "*":
                            result = number1 * number2;
                            break;
                        default:
                            System.out.println("Received unsupported operator: "
                                    + operator);
                            break;
                    }
                    System.out.println("Your expression is:" + number1
                            + operator + number2);
                    System.out.println("Your answer is: " + result);
                } catch (NumberFormatException e) {
                    System.out.println("Invalid expression");

                }
            }
        }
    } catch (FileNotFoundException e) {
        System.out.println("Your file is not found: -" + e.getMessage());
    }

我确信有些东西仍然缺失，因为我完全无法理解你的问题。

明确说明您想要获得的条件。

然后使用上面的代码。

问候。

Answer 3

NumberFormatException的try和catch块没有包装解析字符串为double的正确代码块。

移动try {

try {
   switch (operator) {

要

try {
   Double number1 = Double.parseDouble(inputArray[0]);

保护正确的代码块。

从用户输入

3 个答案: