从父文件连接函数

Question

我有 2 个文件：qdialog1.py 和 qdialog2.py

当qdialog2被拒绝时，我需要从qdialog1连接myfunction

qdialog1 是父级，qdialog2 是其子级

谁能帮帮我？

qdialog1.py

from PyQt4.QtCore import pyqtSignature
from PyQt4.QtGui import QDialog

from Ui_dialog1 import Ui_dialog1
from qdialog2 import Qdialog2

class QDialog1(QDialog, Ui_dialog1):
    def __init__(self, parent=None):
        QDialog.__init__(self, parent)
        self.setupUi(self)

    @pyqtSignature("")
    def on_pbUpdate_clicked(self):
        dlg = QDialog2(self)
        dlg.setModal(True)
        dlg.show()

    def myfunction(self):
        self.lineedit.clear()

qdialog2.py

from PyQt4.QtCore import pyqtSignature
from PyQt4.QtGui import QDialog

from Ui_dialog2 import Ui_dialog2

class QDialog2(QDialog, Ui_dialog2):
    def __init__(self, parent=None):
        QDialog.__init__(self, parent)
        self.setupUi(self)
        self.rejected.connect() # I need help here to call qdialog1.myfunction()

Answer 1

您必须通过父级连接它

self.rejected.connect(parent.myfunction)