将按钮和单独窗口添加到Python QProcess示例

Question

我正在尝试使用QProcess并将stdout读取到由按钮启动的QTextEdit。我如何调整这个例子来做到这一点？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()

Answer 1

使用QPushButton制作按钮。

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()