在Python QProcess示例中添加按钮和单独的窗口

我正在尝试使用QProcess并将标准输出读取到由按钮启动的QTextEdit.我该如何适应this example?我必须为QProcess调用一个单独的类吗?

from PyQt4.QtGui import * 
from PyQt4.QtCore import * 
import sys


class MyQProcess(QProcess):     
  def __init__(self):    
   #Call base class method 
   QProcess.__init__(self)
   #Create an instance variable here (of type QTextEdit)
   self.edit    = QTextEdit()
   self.edit.setWindowTitle("QTextEdit Standard Output Redirection")
   self.edit.show()   

  #Define Slot Here 
  @pyqtSlot()
  def readStdOutput(self):
    self.edit.append(QString(self.readAllStandardOutput()))


def main():  
    app     = QApplication(sys.argv)
    qProcess    = MyQProcess()

    qProcess.setProcessChannelMode(QProcess.MergedChannels);    
    qProcess.start("ldconfig -v")      
    QObject.connect(qProcess,SIGNAL("readyReadStandardOutput()"),qProcess,SLOT("readStdOutput()"));

    return app.exec_()

if __name__ == '__main__':
    main()

解决方法:

使用QPushButton制作一个按钮.

使用QPushButton.clicked.connect绑定事件.

例如:

import sys

from PyQt4.QtGui import *
from PyQt4.QtCore import *


class MyWindow(QWidget):
    def __init__(self):
        QWidget.__init__(self)
        self.edit = QTextEdit()
        self.edit.setWindowTitle("QTextEdit Standard Output Redirection")
        self.button = QPushButton('Run ldconfig')
        self.button.clicked.connect(self.onClick)
        layout = QVBoxLayout(self)
        layout.addWidget(self.edit)
        layout.addWidget(self.button)

    @pyqtSlot()
    def readStdOutput(self):
        self.edit.append(QString(self.proc.readAllStandardOutput()))

    def onClick(self):
        self.proc = QProcess()
        self.proc.start("echo hello")
        self.proc.setProcessChannelMode(QProcess.MergedChannels);
        QObject.connect(self.proc, SIGNAL("readyReadStandardOutput()"), self, SLOT("readStdOutput()"));

def main():
    app = QApplication(sys.argv)
    win = MyWindow()
    win.show()
    return app.exec_()

if __name__ == '__main__':
    main()
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