Qt Python单选按钮:激活事件 [英] Qt Python radiobutton: activate event
问题描述
我正在为一位客户开发一个项目,其中的设计有一个带有独占选项的单选按钮.
I am developing a project for one customer, where the design has a radio button with exclusive options.
这是一段运行并显示两个漂亮单选按钮的代码:
Here is a piece of the code that runs and show two nice radio buttons:
self.performGroupBox = QtGui.QGroupBox(self.centralwidget)
self.performGroupBox.setGeometry(QtCore.QRect(50, 20, 181, 121))
self.performGroupBox.setObjectName("performGroupBox")
self.consultRadioButton = QtGui.QRadioButton(self.performGroupBox)
self.consultRadioButton.setGeometry(QtCore.QRect(40, 30, 84, 18))
self.consultRadioButton.setObjectName("consultRadioButton")
self.insertRadioButton = QtGui.QRadioButton(self.performGroupBox)
self.insertRadioButton.setGeometry(QtCore.QRect(40, 60, 84, 18))
self.insertRadioButton.setObjectName("insertRadioButton")
它看起来像:
perform:
() Consult
() Insert
这里的重点是,如何知道标记了哪个选项:consultRadioButton"还是insertRadioButton"?
The point here is, how to know what choice was marked: "consultRadioButton" or "insertRadioButton"?
以下是尝试获取此信息的示例:
Here is a sample on trying to get this information:
if self.consultRadioButton.isChecked():
self.call_Consult()
if self.insertRadioButton.isChecked():
self.call_Insert()
But it didn't do anything when the radiobutton is chosen.
But it didn't do anything when the radiobutton is chosen.
否则,使用 connect 应该是另一种选择:
Otherwise, using connect should be another option:
QtCore.QObject.connect(self.consultRadioButton, QtCore.SIGNAL("currentIndexChanged(QString)"), self.call_Consult)
QtCore.QObject.connect(self.insertRadioButton, QtCore.SIGNAL("currentIndexChanged(QString)"), self.call_Insert)
但也没有用.
这里缺少什么...有什么建议吗?
What is missing here... Any suggestion?
非常欢迎和感谢所有评论.
All comments are highly welcome and appreciated.
推荐答案
试试这个信号:
void toggled (bool)
https://doc.qt.io/qt-5/qabstractbutton.html#toggle
和示例用法:https://www.tutorialspoint.com/pyqt/pyqt_qradiobutton_widget.htm
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