我正在使用twisted和PyQt在python中编写应用程序。我面临的问题是，当我的扭曲代码中的函数被执行时，我必须在GUI中打印一行，我试图通过发出信号（非PyQt类）来实现这一点。这似乎不起作用，我怀疑扭曲的事件循环正在为PyQt搞砸事情。因为closeEvent信号没有被程序捕获。
以下是代码段：
from PyQt4 import QtGui, QtCore
import sys
from twisted.internet.protocol import Factory, Protocol
from twisted.protocols import amp
import qt4reactor
class register_procedure(amp.Command):
arguments = [('MAC',amp.String()),
('IP',amp.String()),
('Computer_Name',amp.String()),
('OS',amp.String())
]
response = [('req_status', amp.String()),
('ALIGN_FUNCTION', amp.String()),
('ALIGN_Confirmation', amp.Integer()),
('Callback_offset',amp.Integer())
]
class Ui_MainWindow(QtGui.QMainWindow):
def __init__(self,reactor, parent=None):
super(Ui_MainWindow,self).__init__(parent)
self.reactor=reactor
self.pf = Factory()
self.pf.protocol = Protocol
self.reactor.listenTCP(3610, self.pf) # listen on port 1234
def setupUi(self,MainWindow):
MainWindow.setObjectName(_fromUtf8("MainWindow"))
MainWindow.resize(903, 677)
self.centralwidget = QtGui.QWidget(MainWindow)
sizePolicy = QtGui.QSizePolicy(QtGui.QSizePolicy.Expanding, QtGui.QSizePolicy.Expanding)
sizePolicy.setHorizontalStretch(0)
sizePolicy.setVerticalStretch(0)
sizePolicy.setHeightForWidth(self.centralwidget.sizePolicy().hasHeightForWidth())
self.centralwidget.setSizePolicy(sizePolicy)
self.create_item()
self.retranslateUi(MainWindow)
self.connect(self, QtCore.SIGNAL('triggered()'), self.closeEvent)
QtCore.QObject.connect(self,QtCore.SIGNAL('registered()'),self.registered)
QtCore.QMetaObject.connectSlotsByName(MainWindow)
def retranslateUi(self, MainWindow):
MainWindow.setWindowTitle(_translate("MainWindow", "MainWindow", None))
self.pushButton_4.setText(_translate("MainWindow", "Delete System ", None))
self.pushButton.setText(_translate("MainWindow", "Add System", None))
self.label_2.setText(_translate("MainWindow", "SYSTEM STATUS", None))
self.label.setText(_translate("MainWindow", "Monitoring Output", None))
def registered(self):# this function is not being triggered
print "check"
self.textbrowser.append()
def closeEvent(self, event):#neither is this being triggered
print "asdf"
self.rector.stop()
MainWindow.close()
event.accept()
class Protocol(amp.AMP):
@register_procedure.responder
def register_procedure(self,MAC,IP,Computer_Name,OS):
self.bridge_conn=bridge()
cursor_device.execute("""select * FROM devices where MAC = ?;""",[(MAC)])
exists_val=cursor_device.fetchone()
cursor_device.fetchone()
print "register"
if not exists_val== "":
cursor_device.execute("""update devices set IP= ? , Computer_name= ? , OS = ? where MAC= ?;""",[IP,Computer_Name,OS,MAC])
QtCore.QObject.emit( QtCore.SIGNAL('registered')) # <--emits signal
return {'req_status': "done" ,'ALIGN_FUNCTION':'none','ALIGN_Confirmation':0,'Callback_offset':call_offset(1)}
else:
cursor_device.execute("""INSERT INTO devices(Mac,Ip,Computer_name,Os) values (?,?,?,?);""",[MAC,IP,Computer_Name,OS])
QtCore.QObject.emit( QtCore.SIGNAL('registered'))#<--emits signal
return {'req_status': "done" ,'ALIGN_FUNCTION':'main_loop()','ALIGN_Confirmation':0,'Callback_offset':0}
if __name__ == "__main__":
app = QtGui.QApplication(sys.argv)
try:
import qt4reactor
except ImportError:
from twisted.internet import qt4reactor
qt4reactor.install()
from twisted.internet import reactor
MainWindow = QtGui.QMainWindow() # <-- Instantiate QMainWindow object.
ui = Ui_MainWindow(reactor)
ui.setupUi(MainWindow)
MainWindow.show()
reactor.run()
答案 0 :(得分：3)
这是我在我的一个代码中用于从QGraphicsItems发送信号的原因（因为它们不是从QObject派生的，并且默认情况下不能发送/接收信号）。它基本上是Radio-answer的简化版本。
from PyQt4 import QtGui as QG
from PyQt4 import QtCore as QC
class SenderObject(QC.QObject):
something_happened = QC.pyqtSignal()
SenderObject是一个从QObject派生的小类，您可以放置所需的所有信号。在这种情况下，只定义了一个。
class SnapROIItem(QG.QGraphicsRectItem):
def __init__(self, parent = None):
super(SnapROIItem, self).__init__(parent)
self.sender = SenderObject()
def do_something_and_emit(self):
...
self.sender.something_happened.emit()
在非QObject类中，您添加SenderObject作为sender变量。在使用非QObject类的任何地方，您可以将sender的信号连接到您需要的任何信号。
class ROIManager(QC.QObject):
def add_snaproi(self, snaproi):
snaproi.sender.something_happened.connect(...)
<强>更新
完整的代码就是这样，应该打印出“发生了什么......”：
from PyQt4 import QtGui as QG
from PyQt4 import QtCore as QC
class SenderObject(QC.QObject):
something_happened = QC.pyqtSignal()
class SnapROIItem(QG.QGraphicsItem):
def __init__(self, parent = None):
super(SnapROIItem, self).__init__(parent)
self.sender = SenderObject()
def do_something_and_emit(self):
self.sender.something_happened.emit()
class ROIManager(QC.QObject):
def __init__(self, parent=None):
super(ROIManager,self).__init__(parent)
def add_snaproi(self, snaproi):
snaproi.sender.something_happened.connect(self.new_roi)
def new_roi(self):
print 'Something happened in ROI!'
if __name__=="__main__":)
roimanager = ROIManager()
snaproi = SnapROIItem()
roimanager.add_snaproi(snaproi)
snaproi.do_something_and_emit()
更新2
而不是
QtCore.QObject.connect(self,QtCore.SIGNAL('registered()'),self.registered)
你应该：
protocol.sender.registered.connect(self.registered)
这意味着您还需要抓住protocol中的self.pf实例（顺便说一下，您导入Protocol然后自己定义它吗？）
在Protocol类中而不是
QtCore.QObject.emit( QtCore.SIGNAL('registered')
首先，您需要在协议中实例化SenderObject。
class Protocol(amp.AMP):
def __init__( self, *args, **kw ):
super(Protocol, self).__init__(*args, **kw)
self.sender = SenderObject()
然后，在register_procedure中通过sender发出信号：
self.sender.registered.emit（）
要使所有这些工作，您必须将SenderObject定义为：
class SenderObject(QC.QObject):
registered = QC.pyqtSignal()
答案 1 :(得分：1)
这是一篇很老的帖子，但它帮助了我。这是我的版本。一个不是QObject的项目用信号通知另外两个非QObject来运行它们的方法。
from PyQt4 import QtGui, QtCore
class Signal(object):
class Emitter(QtCore.QObject):
registered = QtCore.pyqtSignal()
def __init__(self):
super(Signal.Emitter, self).__init__()
def __init__(self):
self.emitter = Signal.Emitter()
def register(self):
self.emitter.registered.emit()
def connect(self, signal, slot):
signal.emitter.registered.connect(slot)
class item(object):
def __init__(self, name):
self.name = name
self.signal = Signal()
def something(self):
print self.name, ' says something'
>>> itemA = item('a')
>>> itemB = item('b')
>>> itemC = item('c')
>>> itemA.signal.connect(itemA.signal, itemB.something)
>>> itemA.signal.connect(itemA.signal, itemC.something)
>>> itemA.signal.register()
b says something
c says something
答案 2 :(得分：0)
两个基本问题是：
1）必须知道信号的发送者和接收者
在Qt中考虑一个更常见的情况，你可能有多个按钮，每个按钮都有一个“点击”信号。插槽需要知道单击了哪个按钮，因此捕获通用信号没有多大意义。
和2）信号必须来自QObject。
话虽如此，我不确定规范的实现是什么。这是一种方法，使用您在之前的帖子中拥有的Bridge的概念，以及Protocol内部的特殊Emitter类。运行此代码只会在调用protocol.register（）时打印“Working it”。
from PyQt4 import QtGui, QtCore
import sys
class Ui_MainWindow(QtGui.QMainWindow):
def __init__(self):
super(Ui_MainWindow, self).__init__()
def work(self):
print "Working it"
class Protocol(object):
class Emitter(QtCore.QObject):
registered = QtCore.pyqtSignal()
def __init__(self):
super(Protocol.Emitter, self).__init__()
def __init__(self):
self.emitter = Protocol.Emitter()
def register(self):
self.emitter.registered.emit()
class Bridge(QtCore.QObject):
def __init__(self, gui, protocol):
super(Bridge, self).__init__()
self.gui = gui
self.protocol = protocol
def bridge(self):
self.protocol.emitter.registered.connect(self.gui.work)
app = QtGui.QApplication(sys.argv)
gui = Ui_MainWindow()
protocol = Protocol()
bridge = Bridge(gui, protocol)
bridge.bridge()
#protocol.register() #uncomment to see 'Working it' printed to the console