java - 如何用java中的其他字符串替换文件中的字符串

Question

我有一个文件，每一行都包含这样的字符串:

usual,proper,complete,1,convenient,convenient,nonprob,recommended,recommend

我想用这样的代码替换每个单词:

1000, 100, 110, 110, 111, 001, 111, 111, 1000

这是我使用的代码，但它仍然不完整:

public class Codage {
    BufferedReader in;

    public Codage() {
        try {
            in = new BufferedReader(new FileReader("nursery.txt"));
            FileOutputStream fos2 = new FileOutputStream("nursery.txt");
            DataOutputStream output = new DataOutputStream(fos2);
            String str;

            while (null != ((str = in.readLine()))) {

                String delims = ",";
                String[] tokens = str.split(delims);
                int tokenCount = tokens.length;
                for (int j = 0; j < tokenCount; j++) {
                    if (tokens[j].equals("usual")) {
                        tokens[j] = "1000";
                        output.writeChars(tokens[j]);

                    }
                    //continue the other cases
                }
                System.out.print(str);
            }

            in.close();

        } catch (IOException e) {

            System.out.println("There was a problem:" + e);
        }
    }

    public static void main(String[] args) {
        Codage c = new Codage();
    }

}

我的代码错误地替换了值。

Answer 1

首先，您在此处编写的代码不起作用，因为当您打开一个输出流到您尝试读取的确切文件时，它将清空源文件和语句 in.readLine() 总是返回 null。因此，如果这是您的真实代码，那么这可能就是问题所在。

我假设您知道您应该将打开以读取的文件与要写入的文件分开。也就是说，当您打开要读取的 nursery.txt 时，您应该在同一路径中创建一个名为 nursery.tmp 的临时文件的 outputStream，在该过程完成后，您可以删除 nursery.txt 并重命名 nursery .tmp 到 nursery.txt。

另外，如果我是你，我也不会使用 if-else 结构来完成这项工作。它接缝你有独特的 key ，如:

usual, proper, complete, convenient, convenient, nonprob, recommended, recommend

所以也许使用映射结构来查找替换值更方便:

usual, proper, complete, convenient, convenient, nonprob, recommended, recommend, ...

1000, 100, 110, 110, 111, 001, 111, 111, ...

但这些只是一些猜测，您知道如何管理您的业务逻辑。

在那部分之后，我认为将输出数据创建为字符串行并将它们逐行写入 nursery.tmp 是一个更好的主意:

public class Codage {

    private BufferedReader in;
    private BufferedWriter out;

    private HashMap<String, String> replacingValuesByKeys = new HashMap<String, String>();

    public Codage() {
        initialize();
    }

    private void initialize() {
        // I assumed that you have rule that a key like "proper" always goes to "100"
        // Initialize the map between keys and replacing values: 
        replacingValuesByKeys.put("usual", "1000");
        replacingValuesByKeys.put("proper", "100");
        replacingValuesByKeys.put("complete", "110");
        replacingValuesByKeys.put("convenient", "110");
        replacingValuesByKeys.put("nonprob", "111");
        replacingValuesByKeys.put("recommended", "001");
        replacingValuesByKeys.put("recommend", "1000");
    }

    public void doRelpacementInFile(){
        try {
            in = new BufferedReader(new FileReader("c:/nursery.txt"));
            out = new BufferedWriter(new FileWriter("c:/nursery.tmp"));

            String str = in.readLine();
            while (null != str) {
                Iterator<String> it = replacingValuesByKeys.keySet().iterator();
                while(it.hasNext())
                {
                    String toBeReplaced = it.next();
                    String replacementValue = replacingValuesByKeys.get(toBeReplaced);
                    // \\b is for word boundary, because you have both recommend and recommended
                    //        and we do not want to replacing the [recommend] part of recommended.
                    str = str.replaceAll("\\b"+toBeReplaced+"\\b", replacementValue);
                }
                // Write the fully replaced line to the temp file:
                out.append(str);
                out.newLine();

                // Do not forget to read the next line:
                str = in.readLine();
            }

        } catch (IOException e) {
            System.out.println("There was a problem:" + e);
        } finally{
            try {
                in.close();
                out.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }


        File f = new File("c:/nursery.txt");
        f.delete();

        File f2 = new File("c:/nursery.tmp");
        f2.renameTo(new File("c:/nursery.txt"));
    }


    public static void main(String[] args) {
        Codage c = new Codage();
        c.doRelpacementInFile();
    }

}

希望这些片段对您有所帮助，

祝你好运。