python - PyQt4工作线程， "QThread.started"丢失

Question

我正在尝试在 PyQt4 应用程序中正确设置工作线程，但由于某种原因，来自线程的启动信号没有传播到我的工作线程!

    syncThread = QtCore.QThread()
    self._syncThread = syncThread
    worker = SyncWorker(self.async_sync)
    worker.moveToThread(syncThread)
    syncThread.started.connect(self.progress.show) #This dialog appears!
    syncThread.started.connect(worker.work)  # This seems to be a no-op
    worker.finished.connect(syncThread.quit)
    worker.finished.connect(worker.deleteLater)
    syncThread.finished.connect(worker.deleteLater)
    syncThread.finished.connect(syncThread.deleteLater)
    syncThread.start() 

class SyncWorker(QtCore.QObject):

    # Emitted whenever done
    finished = QtCore.pyqtSignal()

    def __init__(self, delegate):
        QtCore.QObject.__init__(self)

    @QtCore.pyqtSlot()
    def work(self):
        print("Worker gonna work") #This never prints!
        self.finished.emit()  

有什么想法吗？

谢谢!

更新:按照 Gary Hughes 重命名worker -> self.worker 后，我在崩溃之前收到一个新错误

QObject::setParent: Cannot set parent, new parent is in a different thread
QPixmap: It is not safe to use pixmaps outside the GUI thread
QPixmap: It is not safe to use pixmaps outside the GUI thread
QPixmap: It is not safe to use pixmaps outside the GUI thread
QPixmap: It is not safe to use pixmaps outside the GUI thread
QPixmap: It is not safe to use pixmaps outside the GUI thread
QPixmap: It is not safe to use pixmaps outside the GUI thread
QPixmap: It is not safe to use pixmaps outside the GUI thread
python: Fatal IO error 11 (Resource temporarily unavailable) on X server :0.
Segmentation fault

更新#2没关系!我的工作人员正在调用 GUI 代码，这导致了新的错误。使用 self.worker 的原始修复是正确的。

Answer 1

尝试将 worker 重命名为 self.worker。

在我看来，worker 在您调用 syncThread.start() 后立即被删除，因为它超出了范围。