java - 如何杀死 watch 服务线程

Question

我需要根据来 self 的 UI 的请求观看 HotFolder我的用户界面上有开始和停止按钮，当我单击开始时，我的代码应该监视该文件夹，单击停止时，它应该停止监视它。我正在使用监视服务来监视 HotFolder，我将标志从 Controller 传递给监视服务以启动和停止监视文件夹。请建议我如何停止观看该文件夹？

下面是代码片段:

@RequestMapping(value = "/start", method = RequestMethod.GET)
public ModelAndView hotFolder()
{
    ModelAndView model = new ModelAndView();
    model.setViewName("welcomePage");

    HotFolder h = new HotFolder();
    h.hotfolderTesting(true);
    return model;
} 

@RequestMapping(value = "/stop", method = RequestMethod.POST)
public ModelAndView hotFolderStop()
{
    ModelAndView model = new ModelAndView();
    model.setViewName("welcomePage");

    HotFolder h = new HotFolder();
    h.hotfolderTesting(false);
    return model;
}

HotFolder.java:

public void hotfolderTesting(boolean flag)
{
    try (WatchService service = FileSystems.getDefault().newWatchService()) {
        Map<WatchKey, Path> keyMap = new HashMap<>();
        Path path = Paths.get("E:\\TestingWatch");
        keyMap.put(path.register(service, StandardWatchEventKinds.ENTRY_CREATE), path);

        WatchKey watchKey;
        if (flag) {
            while (true) {
                watchKey = service.take();
                for (WatchEvent<?> event : watchKey.pollEvents()) {
                    if (event.kind() == StandardWatchEventKinds.ENTRY_CREATE) {
                        System.out.println("Created: " + event.context());
                    } else if (event.kind() == StandardWatchEventKinds.ENTRY_DELETE) {
                        System.out.println("Deleted: " + event.context());
                    } else if (event.kind() == StandardWatchEventKinds.ENTRY_MODIFY) {
                        System.out.println("Modified :" + event.context());
                    }
                }
                if (!watchKey.reset()) {
                    break;
                }
            }
        }
    } catch (Exception ignored) {
    }
}

Answer 1

肯定有一些你没有提到的事情，因为我看到的注释让我认为你正在使用某种基于 REST 的库。

无论如何，你的代码永远不会工作。那是因为你基本上将回答你的问题的线程变成了观看线程，而这永远不会完成你的工作。我建议使用 Singleto 模式，像这样编写 HotFolder 类:

public class HotFolder implements Runnable{
    private static HotFolder instance = null;

    public static HotFolder getInstance(){
        if(instance == null)
            instance = new HotFolder();
        return instance;
    }

    private boolean running = false;
    private Thread t = null
    private HotFolder(){
    }

    public setRunning(boolean running){
        this.running = running;
        if(running && t == null){
            t = new Thread(this);
            t.start()
        }else if(!running && t!= null){
            t = null;
        }
    }

    public boolean getRunning(){
        return running;
    }

    public void run(){
        try (WatchService service = FileSystems.getDefault().newWatchService()) {
            Map<WatchKey, Path> keyMap = new HashMap<>();
            Path path = Paths.get("E:\\TestingWatch");
            keyMap.put(path.register(service, StandardWatchEventKinds.ENTRY_CREATE), path);
            WatchKey watchKey;
            watchKey = service.take();
            do{
                for (WatchEvent<?> event : watchKey.pollEvents()) {
                    if (event.kind() == StandardWatchEventKinds.ENTRY_CREATE) {
                        System.out.println("Created: " + event.context());
                    } else if (event.kind() == StandardWatchEventKinds.ENTRY_DELETE) {
                        System.out.println("Deleted: " + event.context());
                    } else if (event.kind() == StandardWatchEventKinds.ENTRY_MODIFY) {
                        System.out.println("Modified :" + event.context());
                    }
                }
            }while(running && watchKey.reset());
        } catch (Exception ignored) {
        }
    }
}

然后你就这样调用它

//to activate
HotFolder.getInstance().setRunning(true)
//to stop it
HotFolder.getInstance().setRunning(false)