我正在尝试在PyQt中创建视图和控制器,其中视图在单击按钮时发出自定义信号,并且控制器的一个方法连接到发出的信号。但是,它不起作用。单击按钮时不调用响应方法。知道我做错了吗?
import sys
from PyQt4.QtCore import *
from PyQt4.QtGui import QPushButton, QVBoxLayout, QDialog, QApplication
class TestView(QDialog):
def __init__(self, parent=None):
super(TestView, self).__init__(parent)
self.button = QPushButton('Click')
layout = QVBoxLayout()
layout.addWidget(self.button)
self.setLayout(layout)
self.connect(self.button, SIGNAL('clicked()'), self.buttonClicked)
def buttonClicked(self):
self.emit(SIGNAL('request'))
class TestController(QObject):
def __init__(self, view):
self.view = view
self.connect(self.view, SIGNAL('request'), self.respond)
def respond(self):
print 'respond'
app = QApplication(sys.argv)
dialog = TestView()
controller = TestController(dialog)
dialog.show()
app.exec_()
答案 0 :(得分:3)
适合我 - 可能是你正在使用的Qt / PyQt的版本,但有几件事你可以尝试:
您使用的样式是旧式PyQt语法,建议使用新式信号/插槽定义:
import sys
from PyQt4.QtCore import QObject, pyqtSignal # really shouldn't import * here...QtCore library is quite large
from PyQt4.QtGui import QPushButton, QVBoxLayout, QDialog, QApplication
class TestView(QDialog):
request = pyqtSignal()
def __init__(self, parent=None):
super(TestView, self).__init__(parent)
self.button = QPushButton('Click')
layout = QVBoxLayout()
layout.addWidget(self.button)
self.setLayout(layout)
self.button.clicked.connect(self.buttonClicked)
def buttonClicked(self):
self.request.emit()
class TestController(QObject):
def __init__(self, view):
super(QObject, self).__init__()
self.view = view
self.view.request.connect(self.respond)
def respond(self):
print 'respond'
app = QApplication(sys.argv)
dialog = TestView()
controller = TestController(dialog)
dialog.show()
app.exec_()
再说一遍,我真的,真的不鼓励用这种方式构建代码......当你不需要的时候,你正在创造许多不必要的工作和重复的对象。