PyQt对话框在线程运行时不负责

Question

我想通过模态QDialog显示加载进度。所以我创建一个线程来加载数据并在对话框中调用exec()。

loading_progress_dialog = LoadingProgressDialog(len(filenames)) 
loadingWorker = analyzer.LoadingWorker(filenames, loading_progress_dialog.apply_progress) 
workingThread = QThread() 

workingThread.started.connect(loadingWorker.process) 
loadingWorker.finished.connect(workingThread.quit) 
workingThread.finished.connect(loading_progress_dialog.accept) 

loadingWorker.moveToThread(workingThread) 
workingThread.start() 

loading_progress_dialog.exec() 

我想对话框来负责，但它冻结，我无法加载线程运行时，它周围在屏幕上移动。

class LoadingProgressDialog(QLoadingProgressDialog, Ui_LoadingDialog): 
    def __init__(self, maxFiles): 
     super(LoadingProgressDialog, self).__init__() 
     self.setupUi(self) 

     self.progressBar.setMaximum(maxFiles) 
     self.setWindowTitle('Loading files...') 

    def apply_progress(self, delta_progress): 
     self.progressBar.setValue(delta_progress + self.progressBar.value()) 

class LoadingWorker(QtCore.QObject): 
    def __init__(self, file_names, progress_made): 
     super(LoadingWorker, self).__init__() 
     self._file_names = file_names 
     self._progress_made = progress_made 

    finished = QtCore.pyqtSignal() 

    def process(self): 
     print("Thread started") 
     # load_csv_data(self._file_names, self._progress_made)  
     QtCore.QThread.sleep(5) 
     self.finished.emit() 

我与GIL战斗还是另一个问题？我担心的第二件事情是self.finished.emit()和loading_progress_dialog.exec()之间的竞赛状况。如果工作线程完成得比gui线程运行速度快exec()，对话框将不会关闭。有什么方法可以确保一切顺序正确吗？

Answer 1

1. 你的GUI冻结是因为它在同一个线程中的工人执行 - 在主线程！如果您将工作人员移至不同的线程，这会如何呢？好吧，让我们来看看究竟你做了：

   # This connects signal to the instance of worker located in main thread 
   workingThread.started.connect(loadingWorker.process) 
   
   # Creates a copy of worker in the different thread 
   loadingWorker.moveToThread(workingThread) 
   
   # Signal reaches the instance of worker it was connected to - 
   # the instance belonging to main thread! 
   workingThread.start() 
   

   修复的方法是直截了当：招工人以前连接信号给它。如果可以保证进度对话框接收命令关闭一个前显示

2. 赛条件是不可能的：

   class LoadingWorker(QtCore.QObject): 
       [...] 
       def process(self): 
        self.ready.emit() 
        [...] 
        self.finished.emit() 
   
   loadingWorker.ready.connect(loading_progress_dialog.exec) 
   loadingWorker.finished.connect(loading_progress_dialog.close) 
   

因此，简单的程序，它由不同的线程的顺序更新UI可能是这样的：

from PyQt4 import QtGui, QtCore 
from PyQt4.QtCore import QThread 
from time import sleep 

class LoadingProgressDialog(QtGui.QDialog): 
    def __init__(self): 
     super().__init__() 
     self.setWindowTitle('Loading files...') 

    def show_progress(self, p): 
     self.setWindowTitle('Loading files... {}%'.format(p)) 

class LoadingWorker(QtCore.QObject): 
    finished = QtCore.pyqtSignal() 
    ready = QtCore.pyqtSignal() 
    report_progress = QtCore.pyqtSignal(object) 

    def process(self): 
     print('Worker thread ID: %s' % int(QThread.currentThreadId())) 
     print("Worker started") 
     self.ready.emit() 

     for p in range(0, 100, 10): 
      self.report_progress.emit(p) 
      sleep(0.2) 

     print("Worker terminates...") 
     self.finished.emit() 


if __name__ == '__main__': 
    import sys 
    app = QtGui.QApplication([]) 

    print('Main thread ID: %s' % int(QThread.currentThreadId())) 

    workingThread = QThread() 
    loadingWorker = LoadingWorker() 
    loading_progress_dialog = LoadingProgressDialog() 

    loadingWorker.ready.connect(loading_progress_dialog.exec) 
    loadingWorker.report_progress.connect(loading_progress_dialog.show_progress) 
    loadingWorker.finished.connect(workingThread.quit) 
    loadingWorker.finished.connect(loading_progress_dialog.close) 

    loadingWorker.moveToThread(workingThread) 

    workingThread.started.connect(loadingWorker.process) 
    workingThread.start() 

    sys.exit(app.exec_())