Question

我在PyQt中有一个带有函数addImage(image_path)的GUI。很容易想象，当应该将新图像添加到QListWidget中时调用它。为了检测文件夹中的新图像，我使用threading.Thread和watchdog检测文件夹中的文件更改，然后该线程直接调用addImage。

这会产生警告，出于线程安全的原因，不应在gui线程外调用QPixmap。

使线程安全的最佳和最简单的方法是什么？ QThread的？信号/插槽？ QMetaObject.invokeMethod？我只需要将一个字符串从线程传递给addImage。

Answer 1

您应该使用Qt提供的内置QThread。您可以将文件监视代码放在继承自QObject的 worker 类中，以便它可以使用Qt Signal / Slot系统在线程之间传递消息。

class FileMonitor(QObject):

    image_signal = QtCore.pyqtSignal(str)

    @QtCore.pyqtSlot()
    def monitor_images(self):
        # I'm guessing this is an infinite while loop that monitors files
        while True:
            if file_has_changed:
                self.image_signal.emit('/path/to/image/file.jpg')


class MyWidget(QtGui.QWidget):

    def __init__(self, ...)
        ...
        self.file_monitor = FileMonitor()
        self.thread = QtCore.QThread(self)
        self.file_monitor.image_signal.connect(self.image_callback)
        self.file_monitor.moveToThread(self.thread)
        self.thread.started.connect(self.file_monitor.monitor_images)
        self.thread.start()

    @QtCore.pyqtSlot(str)
    def image_callback(self, filepath):
        pixmap = QtGui.QPixmap(filepath)
        ...

Answer 2

我认为最好的方法是使用信号/插槽机制。这是一个例子。（注意：请参阅下面的编辑，指出我的方法可能存在缺陷。）

from PyQt4 import QtGui
from PyQt4 import QtCore

# Create the class 'Communicate'. The instance
# from this class shall be used later on for the
# signal/slot mechanism.

class Communicate(QtCore.QObject):
    myGUI_signal = QtCore.pyqtSignal(str)

''' End class '''


# Define the function 'myThread'. This function is the so-called
# 'target function' when you create and start your new Thread.
# In other words, this is the function that will run in your new thread.
# 'myThread' expects one argument: the callback function name. That should
# be a function inside your GUI.

def myThread(callbackFunc):
    # Setup the signal-slot mechanism.
    mySrc = Communicate()
    mySrc.myGUI_signal.connect(callbackFunc) 

    # Endless loop. You typically want the thread
    # to run forever.
    while(True):
        # Do something useful here.
        msgForGui = 'This is a message to send to the GUI'
        mySrc.myGUI_signal.emit(msgForGui)
        # So now the 'callbackFunc' is called, and is fed with 'msgForGui'
        # as parameter. That is what you want. You just sent a message to
        # your GUI application! - Note: I suppose here that 'callbackFunc'
        # is one of the functions in your GUI.
        # This procedure is thread safe.

    ''' End while '''

''' End myThread '''

在GUI应用程序代码中，您应该创建新的Thread，为其提供正确的回调函数，并使其运行。

from PyQt4 import QtGui
from PyQt4 import QtCore
import sys
import os

# This is the main window from my GUI

class CustomMainWindow(QtGui.QMainWindow):

    def __init__(self):
        super(CustomMainWindow, self).__init__()
        self.setGeometry(300, 300, 2500, 1500)
        self.setWindowTitle("my first window")
        # ...
        self.startTheThread()

    ''''''

    def theCallbackFunc(self, msg):
        print('the thread has sent this message to the GUI:')
        print(msg)
        print('---------')

    ''''''


    def startTheThread(self):
        # Create the new thread. The target function is 'myThread'. The
        # function we created in the beginning.
        t = threading.Thread(name = 'myThread', target = myThread, args = (self.theCallbackFunc))
        t.start()

    ''''''

''' End CustomMainWindow '''


# This is the startup code.

if __name__== '__main__':
    app = QtGui.QApplication(sys.argv)
    QtGui.QApplication.setStyle(QtGui.QStyleFactory.create('Plastique'))
    myGUI = CustomMainWindow()
    sys.exit(app.exec_())

''' End Main '''

修改

先生。 three_pineapples和Brendan Abel先生指出我的方法存在缺陷。实际上，这种方法适用于这种特殊情况，因为您可以直接生成/发出信号。当您处理按钮和小部件上的内置Qt信号时，您应采取另一种方法（如Brendan Abel先生的答案中所述）。

先生。 three_pineapples建议我在StackOverflow中开始一个新主题，以便在与GUI的线程安全通信的几种方法之间进行比较。我会深入研究这个问题，并明天这样做： - ）

PyQT线程最简单的方法

2 个答案: