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是否有C++类的Python版本QtSingleApplication
来自 Qt Solutions ?
QtSingleApplication
用于确保永远不会有多个应用程序实例同时运行。
最佳答案
这是我自己的实现。它已通过 Python 2.7 和 PySide 1.1 测试。
它的界面与C++ version of QtSingleApplication
基本相同。 .主要区别在于您必须向构造函数提供一个应用程序唯一 ID。 (默认情况下,C++ 版本使用可执行文件的路径作为唯一 ID;这在这里不起作用,因为可执行文件很可能是 python.exe
。)
from PySide.QtCore import *
from PySide.QtGui import *
from PySide.QtNetwork import *
class QtSingleApplication(QApplication):
messageReceived = Signal(unicode)
def __init__(self, id, *argv):
super(QtSingleApplication, self).__init__(*argv)
self._id = id
self._activationWindow = None
self._activateOnMessage = False
# Is there another instance running?
self._outSocket = QLocalSocket()
self._outSocket.connectToServer(self._id)
self._isRunning = self._outSocket.waitForConnected()
if self._isRunning:
# Yes, there is.
self._outStream = QTextStream(self._outSocket)
self._outStream.setCodec('UTF-8')
else:
# No, there isn't.
self._outSocket = None
self._outStream = None
self._inSocket = None
self._inStream = None
self._server = QLocalServer()
self._server.listen(self._id)
self._server.newConnection.connect(self._onNewConnection)
def isRunning(self):
return self._isRunning
def id(self):
return self._id
def activationWindow(self):
return self._activationWindow
def setActivationWindow(self, activationWindow, activateOnMessage = True):
self._activationWindow = activationWindow
self._activateOnMessage = activateOnMessage
def activateWindow(self):
if not self._activationWindow:
return
self._activationWindow.setWindowState(
self._activationWindow.windowState() & ~Qt.WindowMinimized)
self._activationWindow.raise_()
self._activationWindow.activateWindow()
def sendMessage(self, msg):
if not self._outStream:
return False
self._outStream << msg << '\n'
self._outStream.flush()
return self._outSocket.waitForBytesWritten()
def _onNewConnection(self):
if self._inSocket:
self._inSocket.readyRead.disconnect(self._onReadyRead)
self._inSocket = self._server.nextPendingConnection()
if not self._inSocket:
return
self._inStream = QTextStream(self._inSocket)
self._inStream.setCodec('UTF-8')
self._inSocket.readyRead.connect(self._onReadyRead)
if self._activateOnMessage:
self.activateWindow()
def _onReadyRead(self):
while True:
msg = self._inStream.readLine()
if not msg: break
self.messageReceived.emit(msg)
这是一个简单的测试程序:
import sys
from PySide.QtGui import *
from QtSingleApplication import QtSingleApplication
appGuid = 'F3FF80BA-BA05-4277-8063-82A6DB9245A2'
app = QtSingleApplication(appGuid, sys.argv)
if app.isRunning(): sys.exit(0)
w = QWidget()
w.show()
app.setActivationWindow(w)
sys.exit(app.exec_())
关于python - PySide 或 PyQt 的 QtSingleApplication,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12712360/
是否有C++类的Python版本QtSingleApplication来自 Qt Solutions ? QtSingleApplication用于确保永远不会有多个应用程序实例同时运行。 最佳答案
自从我使用 QT 5.0.1(来自 http://qt-project.org/downloads)和模块 qtsingleapplication(来自 http://qt.gitorious.org
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