我有几个线程需要与窗口一起使用。这是线程定义:
class MyThread(QtCore.QThread):
def __init__(self, id, window, mutex):
super(MyThread, self).__init__()
self.id = id
self.window = window
self.mutex = mutex
self.connect(self, QtCore.SIGNAL("load_message_input()"), self.window, QtCore.SLOT("show_input()"))
def run(self):
self.mutex.lock()
self.emit(QtCore.SIGNAL("load_message_input()"))
self.connect(self.window, QtCore.SIGNAL("got_message(QString)"), self.print_message)
self.window.input_finished.wait(self.mutex)
self.mutex.unlock()
def print_message(self, str):
print "Thread %d: %s" % (self.id, str)
此处的窗口定义:
class MyDialog(QtGui.QDialog):
def __init__(self, *args, **kwargs):
super(MyDialog, self).__init__(*args, **kwargs)
self.last_message = None
self.setModal(True)
self.message_label = QtGui.QLabel(u"Message")
self.message_input = QtGui.QLineEdit()
self.dialog_buttons = QtGui.QDialogButtonBox(QtGui.QDialogButtonBox.Ok | QtGui.QDialogButtonBox.Cancel)
self.dialog_buttons.accepted.connect(self.accept)
self.dialog_buttons.rejected.connect(self.reject)
self.hbox = QtGui.QHBoxLayout()
self.hbox.addWidget(self.message_label)
self.hbox.addWidget(self.message_input)
self.vbox = QtGui.QVBoxLayout()
self.vbox.addLayout(self.hbox)
self.vbox.addWidget(self.dialog_buttons)
self.setLayout(self.vbox)
self.input_finished = QtCore.QWaitCondition()
@QtCore.pyqtSlot()
def show_input(self):
self.exec_()
def on_accepted(self):
self.emit(QtCore.SIGNAL("got_message(QString)"), self.message_input.text())
self.input_finished.wakeOne()
这是主要的:
if __name__ == "__main__":
import sys
app = QtGui.QApplication(sys.argv)
mutex = QtCore.QMutex()
threads = []
window = test_qdialog.MyDialog()
for i in range(5):
thread = MyThread(i, window, mutex)
thread.start()
threads.append(thread)
for t in threads:
t.wait()
sys.exit(app.exec_())
我无法弄清楚为什么执行脚本时未显示窗口。
更新:由于某些原因,其他线程不会与self.mutex.lock()
保持一致。不知道为什么。
您的代码中有几个问题:
- 如果您希望
QThread
使用插槽,则需要 create 一个事件循环(这很容易,只需致电QThread.exec_
),但是带有事件循环的QThread
S需要进行不同的编码(接下来我会向您发布一个例子) - 如果要发出消息,则需要将
on_accepted
连接到accepted
,除非您使用QT的自动连接功能。 - 如果您要使用
QThread
首先使用QApplication
,因此无法在致电QApplication.exec_
之前执行for t in threads: t.wait()
(在我的示例中刚刚将其删除)。 - 最后一个但同样重要的问题:如果您希望您的线程仅消费资源,则应该考虑一种消费者生产者的方法(问题是,当您发出信号时,每个插槽都会获得数据的副本,以及如果您尝试用事件循环阻止线程,应用程序刚刚冻结,为了解决消费者生产商的问题,我将额外的互音传递给消息的信号,然后 try 锁定它[永远不会阻止它!]知道线程是否消耗事件)
正如承诺的,有一个示例,说明如何在QThread
s上使用事件循环:
from PyQt4 import QtCore, QtGui
class MyThread(QtCore.QThread):
load_message_input = QtCore.pyqtSignal()
def __init__(self, id, window):
super(MyThread, self).__init__()
self.id = id
self.window = window
self.load_message_input.connect(self.window.show_input)
self.window.got_message.connect(self.print_message)
self.started.connect(self.do_stuff)
def run(self):
print "Thread %d: %s" % (self.id,"running")
self.exec_()
@QtCore.pyqtSlot()
def do_stuff(self):
print "Thread %d: %s" % (self.id,"emit load_message_input")
self.load_message_input.emit()
@QtCore.pyqtSlot("QString","QMutex")
def print_message(self, msg, mutex):
if mutex.tryLock():
print "Thread %d: %s" % (self.id, msg)
self.do_stuff()
class MyDialog(QtGui.QDialog):
got_message = QtCore.pyqtSignal("QString","QMutex")
def __init__(self, *args, **kwargs):
super(MyDialog, self).__init__(*args, **kwargs)
self.last_message = None
self.setModal(True)
self.message_label = QtGui.QLabel(u"Message")
self.message_input = QtGui.QLineEdit()
self.dialog_buttons = QtGui.QDialogButtonBox(QtGui.QDialogButtonBox.Ok | QtGui.QDialogButtonBox.Cancel)
self.dialog_buttons.accepted.connect(self.accept)
self.dialog_buttons.accepted.connect(self.on_accepted)
self.dialog_buttons.rejected.connect(self.reject)
self.hbox = QtGui.QHBoxLayout()
self.hbox.addWidget(self.message_label)
self.hbox.addWidget(self.message_input)
self.vbox = QtGui.QVBoxLayout()
self.vbox.addLayout(self.hbox)
self.vbox.addWidget(self.dialog_buttons)
self.setLayout(self.vbox)
self.input_finished = QtCore.QWaitCondition()
@QtCore.pyqtSlot()
def show_input(self):
print "showing input"
window.show()
window.setModal(True)
@QtCore.pyqtSlot()
def on_accepted(self):
print "emit: ", self.message_input.text()
self.got_message.emit(self.message_input.text(), QtCore.QMutex())
if __name__ == "__main__":
import sys
app = QtGui.QApplication(sys.argv)
mutex = QtCore.QMutex()
threads = []
window = MyDialog()
for i in range(5):
thread = MyThread(i, window)
thread.start()
threads.append(thread)
print "start app"
sys.exit(app.exec_())
注意:几乎总是接收信号的线程将是具有ID 1的线。
我的建议,请勿在线程中使用插槽(这将使使用静音和候补条件安全)并为消息实现消费者生产者方法。
您在调用app.exec_()之前正在等待线程退出。您可能应该在GUI空闲循环中监视螺纹,或连接到螺纹的完成()信号。