watchdog在上载zip文件时写入重复文件

Question

下面是通过FileBrowser将文件添加到目录的事件的问题。加载单个文件会产生一堆事件：

我认为消息的数量取决于上传zip文件的时间。我不知道如何用watchdog来解决它，下面是代码：

import time
from watchdog.observers import Observer
from watchdog.events import FileSystemEventHandler
import logging

class Watcher:
    DIRECTORY_TO_WATCH = "/input"

    def __init__(self):
        self.observer = Observer()

    def run(self):
        event_handler = DataHandler()
        self.observer.schedule(event_handler, self.DIRECTORY_TO_WATCH, recursive=False)
        self.observer.start()
        try:
            while True:
                time.sleep(5000)
        except:
            self.observer.stop()
            print("Error")

        self.observer.join()


class DataHandler(FileSystemEventHandler):

    @staticmethod
    def on_any_event(event):
        if event.is_directory:
            return None
        
        elif event.event_type == "created":
            print(f"Received created event {event.src_path}")

        elif event.event_type == "modified":
            print(f"Received modified event {event.src_path}")

if __name__ ==   "__main__":
    w = Watcher()
    w.run()

Answer 1

如果只想为同一文件打印一个事件,可以保存上次修改的路径,仅在新路径不同时打印消息：

class DataHandler(FileSystemEventHandler):
    last_path_modified = ""

    @staticmethod
    def on_any_event(event):
        if event.is_directory:
            return None
        
        elif event.event_type == "created":
            print(f"Received created event {event.src_path}")

        elif event.event_type == "modified":
            if event.src_path != self.last_path_modified:
                print(f"Received modified event {event.src_path}")
                self.last_path_modified = event.src_path

当然，您可以调整这段代码，使其与您想要应用的逻辑相匹配，如果它被标记了x次，则可以再次打印它，或者跟踪第一条消息以来的时间，如果距离上一条消息已超过x秒/分钟，则再次打印它。