watchdog(Python的库) - 如何在修改文件时发送信号?

时间:2016-03-08 17:38:46

标签: python pyqt watchdog python-watchdog

我的代码,用于检测特定文件何时被修改的类:

class MyEventHandler(FileSystemEventHandler, QtCore.QThread):
    def __init__(self, filename):
        super(MyEventHandler, self).__init__()
        self.filename = filename

    def on_modified(self, event):
        if not event.is_directory and event.src_path.endswith(self.filename):
            print "modified"
            self.emit(QtCore.SIGNAL("fileModified"))


class WatchOutForFileModifications(QtCore.QThread):
    def __init__(self, path, filename):
        super(WatchOutForFileModifications, self).__init__()
        self.path = path
        self.filename = filename
        self.observer = Observer()
        self.event_handler = MyEventHandler(self.filename)
        self.observer.schedule(self.event_handler, self.path, recursive=False)
        self.observer.start()

    def run(self):
        while 1:
            self.connect(self.event_handler, QtCore.SIGNAL("fileModified"), self.modified)


    def modified(self):
        self.emit(QtCore.SIGNAL("fileModified1"))

应用程序本身的代码片段:

class MainWindow(QtGui.QMainWindow):
    def __init__(self):
        super(MainWindow, self).__init__()
        uic.loadUi('test.ui', self)

        path = "somePath"
        filename = "someName"

        self.fileWatcher = WatchOutForFileModifications(path, filename)
        self.fileWatcher.start()
        self.connect(self.fileWatcher, QtCore.SIGNAL("fileModified1"), self.fileModified)
        self.show()

    def fileModified(self):
        print 1

问题是,当文件被修改时,我会得到一个不间断的1个正在打印的流。我意识到在WatchOutForFileModifications类中不应该以这种方式发出/连接信号,但我不明白API:http://pythonhosted.org/watchdog/api.html#watchdog.observers.api.EventEmitter - 应该如何工作。至少我认为这是我应该用来监听文件修改的API。

修改

经过一些修改后的工作代码:

导入系统     来自PyQt4导入QtGui,QtCore,uic

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

class MyEventHandler(FileSystemEventHandler, QtCore.QThread):
    def __init__(self, filename):
        super(MyEventHandler, self).__init__()
        self.filename = filename
        self.signalName = str(filename) + "_modified"

    def on_modified(self, event):
        if not event.is_directory and event.src_path.endswith(self.filename):
            self.emit(QtCore.SIGNAL(self.signalName))


class FileModificationWatcher(QtCore.QThread):
    def __init__(self, path, filename):
        super(FileModificationWatcher, self).__init__()
        self.path = path
        self.filename = filename
        self.observer = Observer()
        self.event_handler = MyEventHandler(self.filename)
        self.observer.schedule(self.event_handler, self.path, recursive=False)
        self.observer.start()

    def run(self):
        pass

    def getEmitter(self):
        return self.event_handler

    def getSignalName(self):
        return self.event_handler.signalName

class MainWindow(QtGui.QMainWindow):
    def __init__(self):
        super(MainWindow, self).__init__()
        uic.loadUi('test.ui', self)

        path = "somePath"
        filename = "someName"

        self.fileWatcher = FileModificationWatcher(path, filename)
        self.fileWatcher.start()
        self.connect(self.fileWatcher.getEmitter(), QtCore.SIGNAL(self.fileWatcher.getSignalName()), self.fileModified)
        self.show()

    def fileModified(self):
        print 1

if __name__ == "__main__":
    app = QtGui.QApplication(sys.argv)

    window = MainWindow()
    sys.exit(app.exec_())

2 个答案:

答案 0 :(得分:1)

问题是在WatchOutForFileModifications中你反复将信号连接到运行功能中的插槽。要解决您遇到的问题,请拨打self.connect电话并将其移至课程的__init__,如下所示:

from PyQt4 import QtCore, QtGui
from watchdog.events import FileSystemEventHandler
from watchdog.observers import Observer


class MyEventHandler(FileSystemEventHandler, QtCore.QThread):
    def __init__(self, filename):
        super(MyEventHandler, self).__init__()
        self.filename = filename

    def on_modified(self, event):
        if not event.is_directory and event.src_path.endswith(self.filename):
            print("modified")
            self.emit(QtCore.SIGNAL("fileModified"))


class WatchOutForFileModifications(QtCore.QThread):
    def __init__(self, path, filename):
        super(WatchOutForFileModifications, self).__init__()
        self.path = path
        self.filename = filename
        self.observer = Observer()
        self.event_handler = MyEventHandler(self.filename)
        self.observer.schedule(self.event_handler, self.path, recursive=False)
        self.observer.start()
        self.connect(self.event_handler, QtCore.SIGNAL("fileModified"), self.modified)

    def run(self):
        pass

    def modified(self):
        self.emit(QtCore.SIGNAL("fileModified1"))


class MainWindow(QtGui.QMainWindow):
    def __init__(self):
        super(MainWindow, self).__init__()

        path = r'D:\Code\\'
        filename = "Hexagon_Grid_Creation.py"

        self.fileWatcher = WatchOutForFileModifications(path, filename)
        #self.fileWatcher.start()
        self.connect(self.fileWatcher, QtCore.SIGNAL("fileModified1"), self.fileModified)
        self.show()

    def fileModified(self):
        print(1)

if __name__ == '__main__':
    app = QtGui.QApplication([])
    window = MainWindow()
    window.show()
    app.exec_()

在这种情况下,您可能不需要QThread;事件处理程序正在监视该文件,因此您实际上没有其他任何需要在后台运行的内容。我认为你可以完全取消该类,只需在MainWindow类中实例化事件处理程序。

答案 1 :(得分:-1)

在WatchOutForFileModifications类中,尝试使用 init 方法连接信号。