右键单击QPushButton上的contextMenu

时间:2011-01-27 11:48:40

标签: python qt pyqt contextmenu pyqt4

对于我的应用程序,我在Qt Designer中创建了一个GUI并将其转换为python(2.6)代码。

QPushButton(使用设计器创建)的某些内容中,我想添加一个右键单击上下文菜单。菜单选项取决于应用程序状态。

如何实现这样的上下文菜单?

1 个答案:

答案 0 :(得分:17)

检查以下示例是否适合您。关键是将set context menu policy用于您的小部件到CustomContextMenu并连接到小部件的customContextMenuRequested信号:

import sys
from PyQt4 import QtGui, QtCore

class MainForm(QtGui.QMainWindow):
    def __init__(self, parent=None):
        super(MainForm, self).__init__(parent)

        # create button
        self.button = QtGui.QPushButton("test button", self)       
        self.button.resize(100, 30)

        # set button context menu policy
        self.button.setContextMenuPolicy(QtCore.Qt.CustomContextMenu)
        self.button.customContextMenuRequested.connect(self.on_context_menu)

        # create context menu
        self.popMenu = QtGui.QMenu(self)
        self.popMenu.addAction(QtGui.QAction('test0', self))
        self.popMenu.addAction(QtGui.QAction('test1', self))
        self.popMenu.addSeparator()
        self.popMenu.addAction(QtGui.QAction('test2', self))        

    def on_context_menu(self, point):
        # show context menu
        self.popMenu.exec_(self.button.mapToGlobal(point))        

def main():
    app = QtGui.QApplication(sys.argv)
    form = MainForm()
    form.show()
    app.exec_()

if __name__ == '__main__':
    main()