我是pyqt.I的新手,我正在尝试在单击父GUI中的按钮时调用子GUI。在此过程中,父GUI必须等待用户在选择某些输入后关闭子GUI。但这并没有发生,父GUI确实会执行下一行,在这些行之后调用子GUI。下面是我从父GUI向子GUI传递参数的代码。子GUI将根据单击OK/Cancel按钮返回值
代码:
import sys
from PyQt4 import QtGui,QtCore,Qt
from PyQt4.QtCore import *
from PyQt4.QtGui import *
class Child(QtGui.QWidget):
def __init__(self,switches=None):
super(Child,self).__init__()
self.swwidget = QtGui.QWidget()
self.swlayout = QtGui.QGridLayout()
switches = ['abc1','def1']
switches.sort()
self.switches = switches
def switchesUI(self):
self.swwidget.setWindowModality(QtCore.Qt.ApplicationModal)
self.swl = len(self.switches)
self.sw = {}
self.addsw = []
print ("I am in switchesUI")
#Add the switches to layout dynamically
for i in range(self.swl):
self.sw[i] = QtGui.QCheckBox(self.switches[i])
self.swlayout.addWidget(self.sw[i],i,0)
self.swbuttonbox = QtGui.QDialogButtonBox(QDialogButtonBox.Ok | QDialogButtonBox.Cancel);
self.swbuttonbox.setOrientation(QtCore.Qt.Horizontal)
self.swlayout.addWidget(self.swbuttonbox)
self.swwidget.setWindowTitle('Switches')
self.swwidget.setLayout(self.swlayout)
self.swwidget.show()
self.connect(self.swbuttonbox,QtCore.SIGNAL("accepted()"),self.swaccept)
self.connect(self.swbuttonbox,QtCore.SIGNAL("rejected()"),self.swreject)
def swaccept(self):
for i in range(self.swl):
if self.sw[i].isChecked():
self.addsw.append(self.switches[i])
self.swwidget.close()
return self.addsw
def swreject(self):
self.swwidget.close()
return None
class Parent(QtGui.QWidget):
def __init__(self):
super(Parent,self).__init__()
QtGui.QWidget.__init__(self)
self.button = QtGui.QPushButton('Test', self)
self.layout = QtGui.QVBoxLayout(self)
self.layout.addWidget(self.button)
self.assw = ['Test1','Test2']
self.CH = Child(self.assw)
self.connect(self.button,SIGNAL("clicked()"),self.popup)
print ("Child GUI closed")
def popup(self):
self.CH.switchesUI()
def main():
app = QtGui.QApplication(sys.argv)
form = Parent()
form.show()
sys.exit(app.exec_())
if __name__ == '__main__':
main()单击"Test“按钮后,将弹出一个子GUI。我不希望在关闭子GUI之前打印"Child GUI Closed“语句。有人能建议我如何实现这个功能吗?
发布于 2016-02-18 20:14:00
当窗口要关闭时,您必须处理closeEvent来执行操作,而且由于您的Child类继承自QWidget,这意味着它本身就是一个QWidget,因此您不需要使用self.swwidget创建另一个
import sys
from PyQt4 import QtGui,QtCore,Qt
from PyQt4.QtCore import *
from PyQt4.QtGui import *
class Child(QtGui.QWidget):
def __init__(self,switches=None):
super(Child,self).__init__()
# self.swwidget = QtGui.QWidget() # you don't need to do this you can add all the properties to self
self.swlayout = QtGui.QGridLayout()
switches = ['abc1','def1']
switches.sort()
self.switches = switches
def switchesUI(self):
self.setWindowModality(QtCore.Qt.ApplicationModal)
self.swl = len(self.switches)
self.sw = {}
self.addsw = []
print ("I am in switchesUI")
#Add the switches to layout dynamically
for i in range(self.swl):
self.sw[i] = QtGui.QCheckBox(self.switches[i])
self.swlayout.addWidget(self.sw[i],i,0)
self.swbuttonbox = QtGui.QDialogButtonBox(QDialogButtonBox.Ok | QDialogButtonBox.Cancel);
self.swbuttonbox.setOrientation(QtCore.Qt.Horizontal)
self.swlayout.addWidget(self.swbuttonbox)
self.setWindowTitle('Switches')
self.setLayout(self.swlayout)
self.show()
self.connect(self.swbuttonbox,QtCore.SIGNAL("accepted()"),self.swaccept)
self.connect(self.swbuttonbox,QtCore.SIGNAL("rejected()"),self.swreject)
def swaccept(self):
for i in range(self.swl):
if self.sw[i].isChecked():
self.addsw.append(self.switches[i])
self.close()
return self.addsw
def swreject(self):
self.close()
return None
def closeEvent(self, event):
print ("Child GUI closed")
class Parent(QtGui.QWidget):
def __init__(self):
super(Parent,self).__init__()
QtGui.QWidget.__init__(self)
self.button = QtGui.QPushButton('Test', self)
self.layout = QtGui.QVBoxLayout(self)
self.layout.addWidget(self.button)
self.assw = ['Test1','Test2']
self.CH = Child(self.assw)
self.connect(self.button,SIGNAL("clicked()"),self.popup)
def popup(self):
self.CH.switchesUI()
def main():
app = QtGui.QApplication(sys.argv)
form = Parent()
form.show()
sys.exit(app.exec_())
if __name__ == '__main__':
main()https://stackoverflow.com/questions/35480688
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