当用户尝试再次运行任务栏时，如何最大化或激活打开的应用程序窗口

Question

当用户再次尝试运行应用程序窗口时，如何查看或激活任务栏中存在的应用程序窗口。

例如：我已经打开了MyApp应用程序，它已经存在于任务栏上，并且想要最大化MyApp，如果我试图通过桌面上的应用程序图标或开始菜单再次打开它。

我有这个脚本：

def process():
import psutil
PROCNAME = 'MyApp.exe'
for proc in psutil.process_iter():
try:
if proc.name().lower() == PROCNAME.lower():

return
except (psutil.NoSuchProcess, psutil.AccessDenied, psutil.ZombieProcess):
pass
MainThred = QApplication([])
MainGUI = MainWindow()
MainGUI.show()
sysExit(MainThred.exec_())
if __name__ == "__main__":
process()

Answer 1

您描述了被称为single instance of application的常见模式。您的第二个实例应该知道第一个实例，并向第一个实例发送信号以显示窗口，然后立即退出。因此，您需要某种形式的进程间通信(IPC(。它可以使用tcp套接字或文件句柄或共享内存来实现。

使用共享内存的SingleInstance实现：

from PyQt5.QtCore import QSharedMemory, QTimer, QObject, pyqtSignal
from PyQt5.QtWidgets import QMessageBox, QWidget, QApplication
class SingleInstanceException(Exception):
pass
class SharedMemoryReader(QObject):
received = pyqtSignal(bytes)
def __init__(self, key, parent = None):
super().__init__(parent)
memory = QSharedMemory(key)
if not memory.create(1, QSharedMemory.ReadWrite):
raise SingleInstanceException()

self._memory = memory
memory.lock()
self.clearMemory()
memory.unlock()
# timer to poll shared memory for changes
timer = QTimer()
timer.timeout.connect(self.onTimeOut)
timer.start(500)
self._timer = timer
def clearMemory(self):
self._memory.data()[0] = b'0'

def onTimeOut(self):
memory = self._memory
memory.lock()
data = memory.data()[0]
if data != b'0':
self.received.emit(data)
self.clearMemory()
memory.unlock()
class SharedMemoryWriter(QObject):
def __init__(self, key, parent = None):
super().__init__(parent)
memory = QSharedMemory(key)
self._memory = memory

def write(self, message):
memory = self._memory
if not memory.isAttached():
memory.attach()
memory.lock()
memory.data()[0] = message
memory.unlock()
if __name__ == "__main__":
app = QApplication([])
# guid to identify your application (some unique string)
guid = "5c197822-e86a-4547-a4f1-dbb5087b962d"
widget = QWidget()
try:
# First instance creates shared memory. 
# If memory exists exception raised and it means that it's not first instance.
reader = SharedMemoryReader(guid)
reader.received.connect(lambda: QMessageBox.information(widget,"","Signal from another instance received"))
except SingleInstanceException:
# Second instance writes to memory (sends signal to first instance) and exits
writer = SharedMemoryWriter(guid)
writer.write(b'1')
exit(0)

# No exception raised - that means its first instance of application
# Continue normally
widget.show()
app.exec_()