Question

我正在尝试使用lambda函数连接到一个函数。在这里我想与func CreateInsertLabel（self，qframe）连接这样，每当我单击按钮时，它都会返回一个标签。

这是最小的代码：

from PyQt5 import QtCore, QtGui, QtWidgets

#call this class to create a label
class CreateLabel(QtWidgets.QLabel):
    def __init__(self, qframe,x,y,h,w,font):

        super().__init__(qframe)
        self.state = 2

        self.setFont(font)
        self.move(0,-1000)
        self.setGeometry(x,y,h,w)
        self.setAlignment(QtCore.Qt.AlignCenter)
        self.setObjectName("welcomeMessage")

        self.setText('Insert succeeded!')



class Ui_MainWindow(object):
    def setupUi(self, MainWindow):
        MainWindow.setObjectName("MainWindow")
        MainWindow.resize(800, 600)
        self.centralwidget = QtWidgets.QWidget(MainWindow)
        self.centralwidget.setObjectName("centralwidget")

        self.buttonlabel = QtWidgets.QPushButton(self.centralwidget)
        self.buttonlabel.setGeometry(QtCore.QRect(350, 370, 97, 27))
        self.buttonlabel.setObjectName("buttonlabel")
        #-------------------------------------------
        #The problem here
        self.buttonlabel.clicked.connect(lambda state, x = self.centralwidget: self.CreateInsertLabel(x))

        # This is working as expected
        #self.label_txt = self.CreateInsertLabel(self.centralwidget)

        MainWindow.setCentralWidget(self.centralwidget)
        self.menubar = QtWidgets.QMenuBar(MainWindow)
        self.menubar.setGeometry(QtCore.QRect(0, 0, 800, 25))
        self.menubar.setObjectName("menubar")
        MainWindow.setMenuBar(self.menubar)
        self.statusbar = QtWidgets.QStatusBar(MainWindow)
        self.statusbar.setObjectName("statusbar")
        MainWindow.setStatusBar(self.statusbar)

        self.retranslateUi(MainWindow)
        QtCore.QMetaObject.connectSlotsByName(MainWindow)

    def retranslateUi(self, MainWindow):
        _translate = QtCore.QCoreApplication.translate
        MainWindow.setWindowTitle(_translate("MainWindow", "MainWindow"))
        self.buttonlabel.setText(_translate("MainWindow", "Insert"))


    def CreateInsertLabel(self, qframe):
        font = QtGui.QFont()
        font.setPointSize(15)
        font.setBold(False)
        font.setWeight(20)
        self.label_tour = CreateLabel(qframe, 220, 280, 431, 20,font)


if __name__ == "__main__":
    import sys
    app = QtWidgets.QApplication(sys.argv)
    MainWindow = QtWidgets.QMainWindow()
    ui = Ui_MainWindow()
    ui.setupUi(MainWindow)
    MainWindow.show()
    sys.exit(app.exec_())

如果我使用

可以正常工作

self.label_txt = self.CreateInsertLabel(self.centralwidget)

This is the expected output

在我之前的人有相同的问题（Qt - Connect slot with argument using lambda）和（Using lambda expression to connect slots in pyqt），但这些解决方案对我不起作用。任何帮助将不胜感激。

使用Lambda表达式将插槽与参数连接

0 个答案: