pyqt动态生成QMenu操作并连接 [英] pyqt dynamic generate QMenu action and connect

问题描述

仍在学习pyqt的工作方式.我想动态生成一个customContextMenu并与一个函数连接.到目前为止，我得到了以下内容，但连接部分无法正常工作?

Still learning how pyqt works. I want to dynamically generate a customContextMenu and connect with a function. So far I got the following but the connect part not working ?

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.connect(self.button, QtCore.SIGNAL('customContextMenuRequested(const QPoint&)'), self.on_context_menu)
    self.popMenu = QtGui.QMenu(self)

    def on_context_menu(self, point):
        self.popMenu.clear()

        #some test list for test
        testItems = ['itemA', 'itemB', 'itemC']
        for item in testItems:
            action = self.btn_selectPyFilterPopMenu.addAction("Selected %s" % item)
            self.connect(action,QtCore.SIGNAL("triggered()"),self,QtCore.SLOT("printItem('%s')" % item))    
        self.popMenu.exec_(self.button.mapToGlobal(point))

    @pyqtSlot(str)
    def printItem(self, item):
        print item

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

if __name__ == '__main__':
    main()

推荐答案

您的代码几乎正确.您只需要使用默认参数将信号连接到lambda，如下所示:

Your code is almost right. You just need to connect the signals to a lambda with a default argument, like this:

    for item in testItems:
        action = self.popMenu.addAction('Selected %s' % item)
        action.triggered.connect(
            lambda chk, item=item: self.printItem(item))

默认参数可确保每个lambda获取 current 循环变量的副本.还要注意，还需要一个初始的chk参数.这是因为triggered信号默认发送其当前检查状态(是或否)，这会破坏lambda的item参数.

The default argument ensures that each lambda gets a copy of the current loop variable. Also note that an initial chk argument is also required. This is because the triggered signal sends its current checked-state (true or false) by default, which would clobber the item argument of the lambda.

最后，我强烈建议您在新式语法时使用连接信号-旧样式可能很容易出错，而Python则要少得多.

Finally, I would urge to use the new-style syntax when connecting signals - the old style can be very error-prone, and is far less pythonic.

这篇关于pyqt动态生成QMenu操作并连接的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！