PyQt - 如果UI已经运行,如何检测和关闭它？

Question

我从Maya内部启动UI.如果UI尚未关闭,再次运行UI将完全冻结Maya(错误"事件循环已在运行")

在重新运行脚本之前手动关闭UI将阻止它冻结.但我想这不太实际.

有没有办法检测我正在尝试运行的UI是否已经存在？可能的力量关闭它？

Answer 1

有一对夫妇给予相当简单的C++解决方案在这里.

我已将其中一个移植到PyQt,并在下面提供了一个示例脚本.最初的C++解决方案已分为两类,因为可能不需要消息传递工具.

更新:

改进了脚本,以便它使用新式信号并与python2和python3一起使用.

from PyQt5 import QtCore, QtWidgets

lockfile = QtCore.QLockFile(QtCore.QDir.tempPath() + '/my_app_name.lock')

if lockfile.tryLock(100):
    app = QtWidgets.QApplication([])
    win = QtWidgets.QWidget()
    win.setGeometry(50, 50, 100, 100)
    win.show()
    app.exec()
else:
    print('app is already running')

Answer 2

如果有人想用python3 运行@ekhumoro解决方案,需要对字符串操作进行一些调整,我将分享我的副本,它正在使用python 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_())

示例cli调用,假设您的脚本名称为"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 ?????ó? ?????Ó? etc"

(这里是第一个参数调用,所以不会发送消息)

python SingleInstanceApp.py

希望它会帮助某人.