Question

我想编写Java代码，该代码从文本文件中读取有理数，并逐行添加它们。数字以“和”分隔。但是，总和的行输入错误。

这些是文本文件的内容：

1234/5678and8765/4321
0/1and34/675
apple/23and23/x
-346/74and54/32
-232/884and-33/222
1.2/31and-1/4
-5and1/2
0and3/4
2/3and0
-4/5and5

我已经写了一些代码，但是当输入错误时它会终止。我觉得可以改善

import java.io.*;

class ReadAFile{

    public static void main(String[] args){

        try{

            File myFile = new File("input.txt");
            FileReader fileReader = new FileReader(myFile);

            BufferedReader reader = new BufferedReader(fileReader);

            String line = null;

            while((line=reader.readLine())!=null){

                String [] value = line.split("and");

                String part1 = value[0];
                String part2 = value[1];

                String[] num = part1.split("/");
                String[] dig = part2.split("/");

                float x = Integer.parseInt(num[0]);
                float y = Integer.parseInt(num[1]);

                float a = x/y;

                float p = Integer.parseInt(dig[0]);
                float q = Integer.parseInt(dig[1]);

                float b = p/q;


                float sum = a + b;
                System.out.println(sum);


            }

            reader.close();
        }

        catch(IOException ex){
            ex.printStackTrace();
        }
    }
}

在输出中，我希望在跳过输入错误的行时显示每行具有正确输入的行。

这是我到目前为止的输出：

2.2457957
0.05037037
Exception in thread "main" java.lang.NumberFormatException: For input string: "apple"
        at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
        at java.base/java.lang.Integer.parseInt(Integer.java:652)
        at java.base/java.lang.Integer.parseInt(Integer.java:770)
        at ReadAFile.main(ReadAFile.java:26)

Answer 1

Integer.parseInt引发NumberFormatException，因此您将需要try / catch块。

Answer 2

您可以有效地处理这些行，既可以验证它们，又可以使用单个正则表达式对其进行解析。无需进行所有拆分，因为拆分会产生很多东西。这是一个解决方案：

public static void main(String[] args){

    try {
        File myFile = new File("/tmp/input.txt");
        FileReader fileReader = new FileReader(myFile);

        BufferedReader reader = new BufferedReader(fileReader);

        // The expression for a single float
        String floatPat = "([-+]?[0-9]*\\.?[0-9]+)";

        // Construct expression that models our input...two terms separated by 'and',
        // each of which is a float followed optionally by a slash and a second float.
        String fullPat = "@(/@)?and@(/@)?".replace("@",floatPat);
        Pattern pat = Pattern.compile(fullPat);

        String line = null;
        while((line=reader.readLine())!=null) {

            // Skip empty lines
            if (line.isEmpty())
                continue;

            // apply our expression.  If it matches, process the line
            Matcher m = pat.matcher(line);
            if (m.matches()) {

                // Pull the first value of the first term out of the match
                float x = Float.parseFloat(m.group(1));

                // If the first term had a second part, pull out that float
                float y = (m.group(3) == null)? 1.0f : Float.parseFloat(m.group(3));

                float a = x / y;

                // Pull the first value of the second term out of the match
                float p = Float.parseFloat(m.group(4));

                // If the second term had a second part, pull out that float
                float q = (m.group(6) == null)? 1.0f : Float.parseFloat(m.group(6));

                float b = p / q;

                float sum = a + b;
                System.out.println(sum);
            }
            else {
                // Do something with  bad input
                System.out.println("Bad input: " + line);
            }
        }

        reader.close();
    }
    catch (IOException ex){
        ex.printStackTrace();
    }
}

请注意，在这里，我也选择打印无效行，并注明。

输出：

2.2457957
0.05037037
Bad input: apple/23and23/x
-2.9881759
-0.4110921
-0.21129033
-4.5
0.75
0.6666667
4.2

Answer 3

使用 regex（“ [0-9 /.-]+")来验证零件数据是否仅包含有理数。您必须使用Float.parseFloat（）代替parseInt（）。 parseInt会为浮点数抛出NumberFormatException。

例如：：在解析文件中的第6行（Integer.parseInt（“ 1.2”）;）时，将引发NumberFormatException。

package com.stackovflow.problems;

import java.io.*;

class ReadAFile {

    public static void main(String[] args) {

        try {

            File myFile = new File("input.txt");
            FileReader fileReader = new FileReader(myFile);

            BufferedReader reader = new BufferedReader(fileReader);

            String line = null;
            String onlyFloatNumRegex = "-?[0-9.]+";
            while ((line = reader.readLine()) != null) {

                String[] value = line.split("and");
                if (value.length == 2) {
                    String part1 = value[0];
                    String part2 = value[1];
                    if (part1.matches("[0-9/.-]+") && part2.matches("[0-9/.-]+")) {

                        String[] num = part1.split("/");
                        String[] dig = part2.split("/");
                        if (num.length == 2 && dig.length == 2 && num[0].matches(onlyFloatNumRegex) && num[1].matches(onlyFloatNumRegex)
                                && dig[0].matches(onlyFloatNumRegex) && dig[1].matches(onlyFloatNumRegex)) {
                            float x = Float.parseFloat(num[0]);
                            float y = Float.parseFloat(num[1]);

                            float a = x / y;

                            float p = Float.parseFloat(dig[0]);
                            float q = Float.parseFloat(dig[1]);

                            float b = p / q;

                            float sum = a + b;
                            System.out.println(sum);

                        }

                    }
                }

            }

            reader.close();
        }

        catch (IOException ex) {
            ex.printStackTrace();
        }
    }
}

从文本文件中逐行读取有理数，并在每行中添加数字，其中某些行输入错误

3 个答案: