2012-01-09 59 views
7

我從Maya開始使用UI。如果UI尚未關閉,再次運行UI將完全凍結Maya(錯誤「事件循環已在運行」)PyQt - 如何檢測和關閉用戶界面,如果它已經運行?

在重新運行腳本之前手動關閉UI將防止它凍結。但我想這不是很實際。

有沒有辦法檢測我試圖運行的UI是否已經存在?可能的力量關閉它?

+0

可能是重複的http://stackoverflow.com/questions/5006547/qt-best-practice-for-a-single-instance-app-protection – 2012-01-09 09:26:56

回答

15

有幾個相當簡單的C++解決方案給出here

我已將其中一個移植到PyQt,並提供了下面的示例腳本。最初的C++解決方案已被分爲兩類,因爲可能不需要消息傳遞工具。

UPDATE

改進腳本,以便它使用新型信號,既python2和python3工作。

# only needed for python2 
import sip 
sip.setapi('QString', 2) 

from PyQt4 import QtGui, QtCore, QtNetwork 

class SingleApplication(QtGui.QApplication): 
    messageAvailable = QtCore.pyqtSignal(object) 

    def __init__(self, argv, key): 
     QtGui.QApplication.__init__(self, argv) 
     self._memory = QtCore.QSharedMemory(self) 
     self._memory.setKey(key) 
     if self._memory.attach(): 
      self._running = True 
     else: 
      self._running = False 
      if not self._memory.create(1): 
       raise RuntimeError(self._memory.errorString()) 

    def isRunning(self): 
     return self._running 

class SingleApplicationWithMessaging(SingleApplication): 
    def __init__(self, argv, key): 
     SingleApplication.__init__(self, argv, key) 
     self._key = key 
     self._timeout = 1000 
     self._server = QtNetwork.QLocalServer(self) 
     if not self.isRunning(): 
      self._server.newConnection.connect(self.handleMessage) 
      self._server.listen(self._key) 

    def handleMessage(self): 
     socket = self._server.nextPendingConnection() 
     if socket.waitForReadyRead(self._timeout): 
      self.messageAvailable.emit(
       socket.readAll().data().decode('utf-8')) 
      socket.disconnectFromServer() 
     else: 
      QtCore.qDebug(socket.errorString()) 

    def sendMessage(self, message): 
     if self.isRunning(): 
      socket = QtNetwork.QLocalSocket(self) 
      socket.connectToServer(self._key, QtCore.QIODevice.WriteOnly) 
      if not socket.waitForConnected(self._timeout): 
       print(socket.errorString()) 
       return False 
      if not isinstance(message, bytes): 
       message = message.encode('utf-8') 
      socket.write(message) 
      if not socket.waitForBytesWritten(self._timeout): 
       print(socket.errorString()) 
       return False 
      socket.disconnectFromServer() 
      return True 
     return False 

class Window(QtGui.QWidget): 
    def __init__(self): 
     QtGui.QWidget.__init__(self) 
     self.edit = QtGui.QLineEdit(self) 
     self.edit.setMinimumWidth(300) 
     layout = QtGui.QVBoxLayout(self) 
     layout.addWidget(self.edit) 

    def handleMessage(self, message): 
     self.edit.setText(message) 

if __name__ == '__main__': 

    import sys 

    key = 'app-name' 

    # send commandline args as message 
    if len(sys.argv) > 1: 
     app = SingleApplicationWithMessaging(sys.argv, key) 
     if app.isRunning(): 
      print('app is already running') 
      app.sendMessage(' '.join(sys.argv[1:])) 
      sys.exit(1) 
    else: 
     app = SingleApplication(sys.argv, key) 
     if app.isRunning(): 
      print('app is already running') 
      sys.exit(1) 

    window = Window() 
    app.messageAvailable.connect(window.handleMessage) 
    window.show() 

    sys.exit(app.exec_()) 
+0

如果我運行與消息傳遞我總是得到這個錯誤:'QLocalSocket :: connectToServer:連接拒絕'任何想法如何解決它? – Jeena 2013-04-25 22:38:13

+0

@Jeena。用PyQt-4.10在Linux上仍然工作。 – ekhumoro 2013-04-26 00:48:06

+0

哦,那只是一些隨機的怪事,今晚我的電腦重啓後它似乎現在工作得很好,謝謝你回來了! – Jeena 2013-04-26 20:05:42

8

萬一,如果有人想運行@ekhumoro解決方案與python3有需要作出一些調整,以字符串操作,我會分享我的副本,它是工作蟒蛇3

import sys 

from PyQt4 import QtGui, QtCore, QtNetwork 

class SingleApplication(QtGui.QApplication): 
    def __init__(self, argv, key): 
     QtGui.QApplication.__init__(self, argv) 
     self._memory = QtCore.QSharedMemory(self) 
     self._memory.setKey(key) 
     if self._memory.attach(): 
      self._running = True 
     else: 
      self._running = False 
      if not self._memory.create(1): 
       raise RuntimeError(self._memory.errorString()) 

    def isRunning(self): 
     return self._running 

class SingleApplicationWithMessaging(SingleApplication): 
    def __init__(self, argv, key): 
     SingleApplication.__init__(self, argv, key) 
     self._key = key 
     self._timeout = 1000 
     self._server = QtNetwork.QLocalServer(self) 

     if not self.isRunning(): 
      self._server.newConnection.connect(self.handleMessage) 
      self._server.listen(self._key) 

    def handleMessage(self): 
     socket = self._server.nextPendingConnection() 
     if socket.waitForReadyRead(self._timeout): 
      self.emit(QtCore.SIGNAL('messageAvailable'), bytes(socket.readAll().data()).decode('utf-8')) 
      socket.disconnectFromServer() 
     else: 
      QtCore.qDebug(socket.errorString()) 

    def sendMessage(self, message): 
     if self.isRunning(): 
      socket = QtNetwork.QLocalSocket(self) 
      socket.connectToServer(self._key, QtCore.QIODevice.WriteOnly) 
      if not socket.waitForConnected(self._timeout): 
       print(socket.errorString()) 
       return False 
      socket.write(str(message).encode('utf-8')) 
      if not socket.waitForBytesWritten(self._timeout): 
       print(socket.errorString()) 
       return False 
      socket.disconnectFromServer() 
      return True 
     return False 

class Window(QtGui.QWidget): 
    def __init__(self): 
     QtGui.QWidget.__init__(self) 
     self.edit = QtGui.QLineEdit(self) 
     self.edit.setMinimumWidth(300) 
     layout = QtGui.QVBoxLayout(self) 
     layout.addWidget(self.edit) 

    def handleMessage(self, message): 
     self.edit.setText(message) 

if __name__ == '__main__': 

    key = 'foobar' 

    # if parameter no. 1 was set then we'll use messaging between app instances 
    if len(sys.argv) > 1: 
     app = SingleApplicationWithMessaging(sys.argv, key) 
     if app.isRunning(): 
      msg = '' 
      # checking if custom message was passed as cli argument 
      if len(sys.argv) > 2: 
       msg = sys.argv[2] 
      else: 
       msg = 'APP ALREADY RUNNING' 
      app.sendMessage(msg) 
      print("app is already running, sent following message: \n\"{0}\"".format(msg)) 
      sys.exit(1) 
    else: 
     app = SingleApplication(sys.argv, key) 
     if app.isRunning(): 
      print('app is already running, no message has been sent') 
      sys.exit(1) 

    window = Window() 
    app.connect(app, QtCore.SIGNAL('messageAvailable'), window.handleMessage) 
    window.show() 

    sys.exit(app.exec_()) 

例cli來電話,假設你的腳本名稱是 「SingleInstanceApp.py」:

python SingleInstanceApp.py 1 
python SingleInstanceApp.py 1 "test" 
python SingleInstanceApp.py 1 "foo bar baz" 
python SingleInstanceApp.py 1 "utf8 test FOO ßÄÖÜ ßäöü łąćźżóń ŁĄĆŹŻÓŃ etc" 

(這裏是電話wihout第一個參數,這樣的消息根本不會被髮送)

python SingleInstanceApp.py

希望它能幫助別人。

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