如何在分离模式下终止进程

时间:2020-02-05 14:30:14

标签: python pyqt pyqt5 qprocess

我使用PyQt5在Python 3.8中开发应用。

我需要运行并能够杀死与PyQT表单不同的进程。 当我使用 process.start 然后使用 process.kill 时,它会终止该进程。

但是当我使用process.startDetached()时,process.kill()不起作用。

下面是一个示例:

from form import Ui_MainWindow
from PyQt5 import QtWidgets

class myapp(QtWidgets.QMainWindow):

    def __init__(self):
        super(myapp, self).__init__()
        self.ui = Ui_MainWindow()
        self.ui.setupUi(self)
        self.process = QProcess(self)
        self.ui.pushButton.clicked.connect(self.btnClicked)
        self.ui.pushButton_2.clicked.connect(self.process.kill)

    def btnClicked(self):
        runstr = 'ping'
        args = ['localhost','-t']
        self.process.startDetached(runstr, args)

app = QtWidgets.QApplication([])
application = myapp()
application.show()
sys.exit(app.exec())

您能否建议如何以分离模式杀死进程。

1 个答案:

答案 0 :(得分:0)

启动分离的进程时,QProcess无法终止该进程,因此在这些情况下,您必须获取子进程的pid,并使用OS方法将其终止,如以下问题的答案所示:

from PyQt5 import QtCore, QtWidgets

import psutil

from form import Ui_MainWindow


class MyApp(QtWidgets.QMainWindow):
    def __init__(self, parent=None):
        super(MyApp, self).__init__(parent)
        self.ui = Ui_MainWindow()
        self.ui.setupUi(self)
        self.process = QtCore.QProcess(self)
        self.ui.pushButton.clicked.connect(self.start_process)
        self.ui.pushButton_2.clicked.connect(self.stop_process)
        self._pid = -1

    @QtCore.pyqtSlot()
    def start_process(self):
        runstr = "ping"
        args = ["localhost", "-t"]
        self.process.setProgram(runstr)
        self.process.setArguments(args)
        ok, pid = self.process.startDetached()
        if ok:
            self._pid = pid

    @QtCore.pyqtSlot()
    def stop_process(self):
        if self._pid > 0:
            p = psutil.Process(self._pid)
            p.terminate()
            self._pid = -1


if __name__ == "__main__":

    import sys

    app = QtWidgets.QApplication(sys.argv)
    application = MyApp()
    application.show()
    sys.exit(app.exec_())