Java 7 WatchService - 进程无法访问该文件,因为它正由另一个进程使用

Question

我按照观察变更目录 Java7 nio2教程,使用代码示例WatchDir.java以递归方式监视目录的全部内容.

代码如下所示:

// Get list of events for the watch key.
for (WatchEvent<?> event : key.pollEvents()) {
// This key is registered only for ENTRY_CREATE events, but an OVERFLOW event 
// can occur regardless if events are lost or discarded.
if (event.kind() == OVERFLOW) {
    continue;
}

// Context for directory entry event is the file name of entry.
@SuppressWarnings("unchecked")
WatchEvent<Path> ev = (WatchEvent<Path>)event;
Path fileName = ev.context();
Path fullPath = dir.resolve(fileName);

try {
    // Print out event.
    System.out.print("Processing file: " + fileName);

    processed = fileProcessor.processFile(fullPath);

    System.out.println("Processed = " + processed);

    if (processed) {
        // Print out event.
        System.out.println(" - Done!");
    }
} 
catch (FileNotFoundException e) {
    System.err.println("Error message: " + e.getMessage());
}
catch (IOException e) {
    System.err.println("Error processing file: " + fileName.toString());
    System.err.println("Error message: " + e.getMessage());
}

好的,所以问题(我肯定做一些愚蠢的事情)就在这里:

processed = fileProcessor.processFile(fullPath);

它的作用是这样的:

public synchronized boolean processFile(Path fullPath) throws IOException {
String line;
String[] tokens;
String fileName = fullPath.getFileName().toString();
String fullPathFileName = fullPath.toString();

// Create the file.
File sourceFile = new File(fullPath.toString());

// If the file does not exist, print out an error message and return.
if (sourceFile.exists() == false) {
    System.err.println("ERROR: " + fullPathFileName + ": No such file");
    return false;
}

// Check file extension.
if (!getFileExtension(fullPathFileName).equalsIgnoreCase("dat")) {
    System.out.println(" - Ignored.");
    return false;
}

// Process source file.
try (BufferedReader bReader = new BufferedReader(new FileReader(sourceFile))) {
    int type;

    // Process each line of the file.
    while (bReader.ready()) {

        // Get a single line.
        line = bReader.readLine();

        // Get line tokens.
        tokens = line.split(delimiter);

        // Get type.
        type = Integer.parseInt(tokens[0]);

        switch (type) {
        // Type 1 = Salesman.
        case 1:
            -> Call static method to process tokes.
            break;
        // Type 2 = Customer.
        case 2:
            -> Call static method to process tokes.
            break;
        // Type 3 = Sales.
        case 3:
            -> Call static method to process tokes.
            break;
        // Other types are unknown!
        default:
            System.err.println("Unknown type: " + type);
            break;
        }
    }

    PrintStream ps = null;
    try {

        // Write output file.
        // Doesn't matter. 

    } 
    finally {               
        if (ps != null) {
            ps.close();
        }
    }
    return true;
}
}

我第一次处理一个事件,一切正常!即使有多个文件要处理.但在连续的时间里,我收到此错误消息:

该进程无法访问该文件,因为该文件正由另一个进程使用

我在这做错了什么？如何通过成功处理连续文件？

我忘了提到两个重要的注意事项:

1. 我正在使用Windows 7.
2. 当我在调试模式下运行应用程序时,它可以工作.

编辑:如果我在尝试使用该文件之前添加睡眠,它的工作原理如下:

Thread.sleep(500);

// Process source file.
try (BufferedReader bReader = new BufferedReader(new FileReader(sourceFile))) {

那么,Windows是否可能无法及时解锁文件？我该如何解决(以适当的方式)？

Answer 1

好的,我找到了解决方案.我不知道这是否是最好的方法,但它确实有效.不幸的是,即使文件仍被Windows锁定,file.canRead()和file.canWrite()都返回true.所以我发现如果我尝试用同一个名称"重命名"它,我知道Windows是否正在使用它.所以这就是我做的:

    while(!sourceFile.renameTo(sourceFile)) {
        // Cannot read from file, windows still working on it.
        Thread.sleep(10);
    }