PyQt4信号和插槽eventFilter

Question

我无法接收事件过滤器发送的信号。在下面的例子中，按下按钮的信号/插槽工作正常，并且焦点输出过滤器信号发出OK。然而，焦点输出信号不会被拦截，插槽也不会触发。任何想法我做错了什么？

from PyQt4.QtCore import SIGNAL, QObject, QEvent 
from PyQt4.QtGui import QApplication, QLabel, QWidget, QLineEdit, QPushButton, QTextEdit, QVBoxLayout 

class SignalOnFocus(QWidget): 
    def __init__(self): 
     super(SignalOnFocus, self).__init__() 
     layout = QVBoxLayout() 
     self.label = QLabel("Type in some text then push button") 
     self.inputLineEdit1 = QLineEdit() 
     self.inputLineEdit1.setObjectName("inputLineEdit1") 

     self.focusOutFilter = FocusOutFilter() 
     self.inputLineEdit1.installEventFilter(self.focusOutFilter) 
     self.connect(self.inputLineEdit1, SIGNAL("focus_out"), 
       self.focusLost) 
     self.inputLineEdit2 = QLineEdit() 
     self.inputLineEdit2.setObjectName("inputLineEdit2") 
     self.mousePressedFilter = MousePressedFilter() 
     self.inputLineEdit2.installEventFilter(self.mousePressedFilter) 
     self.connect(self.inputLineEdit2, SIGNAL("mouse_clicked"), self.mouseClicked) 
     self.button1 = QPushButton("Press me") 
     self.button1.setObjectName("button1") 
     self.connect(self.button1, SIGNAL("clicked()"), self.buttonPressed) 
     self.textEdit = QTextEdit() 
     layout.addWidget(self.label) 
     layout.addWidget(self.inputLineEdit1) 
     layout.addWidget(self.inputLineEdit2) 
     layout.addWidget(self.button1) 
     layout.addWidget(self.textEdit) 
     self.setLayout(layout) 

    def mouseClicked(self): 
     self.textEdit.append(" mouse clicked") 

    def buttonPressed(self): 
     self.textEdit.append(" button pressed") 

    def focusLost(self): 
     self.textEdit.append(" focus_out") 

class MousePressedFilter(QObject): 
    def eventFilter(self, widget, event): 
     if event.type() == QEvent.MouseButtonPress: 
      print("--eventFilter() mouse_clicked on "+str(widget.objectName())) 
      self.emit(SIGNAL("mouse_clicked")) 
      return False 
     else: 
      return False 

class FocusOutFilter(QObject): 
    def eventFilter(self, widget, event): 
     if event.type() == QEvent.FocusOut: 
      print("--eventFilter() focus_out on "+str(widget.objectName())) 
      self.emit(SIGNAL("focus_out")) 
      return False 
     else: 
      return False 

if __name__ == "__main__": 
    app = QApplication([]) 
    form = SignalOnFocus() 
    form.show() 
    app.exec_() 

Answer 1

过滤对象发出信号，所以这是你所需要的指定时，将它们连接起来：

self.connect(self.focusOutFilter, SIGNAL("focus_out"), self.focusLost) 
    ... 
    self.connect(self.mousePressedFilter, SIGNAL("mouse_clicked"), self.mouseClicked) 

但是请认真考虑连接摆脱那个丑陋的，旧式的语法信号。官方对Qt4的支持即将在今年结束，而PyQt5已经使旧式语法完全过时了。

使用new-style syntax，你的例子是这样的：

from PyQt4.QtCore import pyqtSignal, QObject, QEvent 

class SignalOnFocus(QWidget): 
    def __init__(self): 
     ...  
     self.focusOutFilter = FocusOutFilter() 
     self.inputLineEdit1.installEventFilter(self.focusOutFilter) 
     self.focusOutFilter.focusOut.connect(self.focusLost) 

class FocusOutFilter(QObject): 
    focusOut = pyqtSignal() 

    def eventFilter(self, widget, event): 
     if event.type() == QEvent.FocusOut: 
      print("--eventFilter() focus_out on " + widget.objectName()) 
      self.focusOut.emit() 

，我希望你会同意看上去更具可读性（而且更容易得到正确的）。 （也请注意，如果您使用Python 3与PyQt，默认情况下，任何返回QString的Qt方法是automatically converted to a python string - 因此您不需要使用str自己转换它）。