Python 用于获取标准输出的 PyQt5 信号不起作用

Question

我遇到了一个很奇怪的错误。我正在尝试将 QProcess 输出重定向到 textbrowser.append() 在调试模式下，它在 VS 和 VSCode 中工作，但在发布模式下不起作用。信号 readyReadStandardError 和 readyReadStandardOut 在此过程中不发出信号。只有当它完成时。但是我需要它在这个过程中发出信号。

我已经试过了

寻找通道中的问题（setProcessChannelMode（QtCore.QProcess.ForwardedOutputChannel）正在工作并实时打印到 main.sys。stdout 但我只需要 stdout从 Qprocess 传输到 QtextBrowser。当我将频道切换到 SeparateChannels 或 MergedChannels 时，我没有输出。
尝试了不同的信号方法（-started()、-errorOccured()）

我的代码看起来像这样：

    self.process = QtCore.QProcess()
    self.process.setProgram("python")
    self.process.setArguments(["run.py"])

    #both not working when no debugging mode in VSCode
    #self.process.readyReadStandardError.connect(self.on_readyReadStandardError)
    #self.process.readyReadStandardOutput.connect(self.on_readyReadStandardOutput)

    #ForwardedOutputChannel is working
    self.process.setProcessChannelMode(QtCore.QProcess.ForwardedOutputChannel)

    #another try not working
    #self.process.setReadChannel(QtCore.QProcess.StandardOutput)

    #tried here a different signal with no luck
    #self.process.started.connect(self.on_readyReadStandardOutput)
    
    self.pushButton.clicked.connect(self.accepted)

  #starting process on button click
  def accepted(self):
    self.process.start()

  #what to do when readyReadStandardOutput signal is emitted - i did the same for stderr
  def on_readyReadStandardOutput(self):
    print("hi") #test if the signal is being send
    out = self.processSubscribe.readAllStandardOutput().data().decode().strip()
    self.textBrowser.append(out)

编辑这里有一小段测试：

from PyQt5 import QtCore, QtGui, QtWidgets


class Ui_MainWindow(object):
    def setupUi(self, MainWindow):
        MainWindow.setObjectName("MainWindow")
        MainWindow.resize(800, 600)
        self.centralwidget = QtWidgets.QWidget(MainWindow)
        self.centralwidget.setObjectName("centralwidget")
        self.gridLayout = QtWidgets.QGridLayout(self.centralwidget)
        self.gridLayout.setObjectName("gridLayout")
        self.pushButton = QtWidgets.QPushButton(self.centralwidget)
        self.pushButton.setObjectName("pushButton")
        self.gridLayout.addWidget(self.pushButton, 0, 0, 1, 1)
        self.textBrowser = QtWidgets.QTextBrowser(self.centralwidget)
        self.textBrowser.setObjectName("textBrowser")
        self.gridLayout.addWidget(self.textBrowser, 1, 0, 1, 1)
        MainWindow.setCentralWidget(self.centralwidget)
        self.menubar = QtWidgets.QMenuBar(MainWindow)
        self.menubar.setGeometry(QtCore.QRect(0, 0, 800, 22))
        self.menubar.setObjectName("menubar")
        MainWindow.setMenuBar(self.menubar)
        self.statusbar = QtWidgets.QStatusBar(MainWindow)
        self.statusbar.setObjectName("statusbar")
        MainWindow.setStatusBar(self.statusbar)

        self.retranslateUi(MainWindow)
        QtCore.QMetaObject.connectSlotsByName(MainWindow)

    def retranslateUi(self, MainWindow):
        _translate = QtCore.QCoreApplication.translate
        MainWindow.setWindowTitle(_translate("MainWindow", "MainWindow"))
        self.pushButton.setText(_translate("MainWindow", "PushButton"))

        self.process = QtCore.QProcess()
        self.process.setProgram("python")
        self.process.setArguments(["[INPUT PATH TO FILE]\run.py"])

        #both not working when no debugging mode in VSCode
        self.process.readyReadStandardError.connect(self.on_readyReadStandardError)
        self.process.readyReadStandardOutput.connect(self.on_readyReadStandardOutput)

        #ForwardedOutputChannel is working
        #self.process.setProcessChannelMode(QtCore.QProcess.ForwardedOutputChannel)

        self.pushButton.clicked.connect(self.accepted)

    #starting process on button click
    def accepted(self):
        self.process.start()

  #what to do when readyReadStandardOutput signal is emitted - i did the same for stderr
    def on_readyReadStandardOutput(self):
        print("hi") #test if the signal is being send
        out = self.process.readAllStandardOutput().data().decode().strip()
        self.textBrowser.append(out)

    def on_readyReadStandardError(self):
        err = self.process.readAllStandardError().data().decode().strip()
        self.textBrowser.append(err)


if __name__ == "__main__":
    import sys
    app = QtWidgets.QApplication(sys.argv)
    MainWindow = QtWidgets.QMainWindow()
    ui = Ui_MainWindow()
    ui.setupUi(MainWindow)
    MainWindow.show()
    sys.exit(app.exec_())

run.py 看起来像这样（QProcess 的外部 python 程序）

import time
for i in range(5):
    print(i)
    time.sleep(1)

自己复制试试

Answer 1

我找到了正确的答案。我编辑 run.py 行

sys.stdout.flush()

所以：

import time
for i in range(5):
  print(i)
  sys.stdout.flush()
  time.sleep(1)

Python 用于获取标准输出的 PyQt5 信号不起作用

Python PyQt5 signal for getting stdout not working

python

stdout

qprocess

pyqt5