下面的代码在self.emit行上中断。它在PyQt4中工作正常。如何修复此代码,使其在PyQt5中正常工作?

from PyQt5 import QtCore, QtGui, QtWidgets
from PyQt5.QtCore import QObject, pyqtSignal

class ItemDelegate(QtWidgets.QItemDelegate):
    def __init__(self, parent):
        QtWidgets.QItemDelegate.__init__(self, parent)

    def createEditor(self, parent, option, index):
        return QtWidgets.QLineEdit()

    @QtCore.pyqtSlot()
    def setModelData(self, editor, model, index):
        self.emit(QtCore.SIGNAL("dataChanged(QModelIndex,QModelIndex)"), index, index)

以后编辑:

一个可行的解决方案:
from PyQt5 import QtCore, QtGui, QtWidgets
from PyQt5.QtCore import QObject, pyqtSignal

class Communicate(QObject):
    data_changed = pyqtSignal(QtCore.QModelIndex, QtCore.QModelIndex)

class ItemDelegate(QtWidgets.QItemDelegate):
    def __init__(self, parent):
        QtWidgets.QItemDelegate.__init__(self, parent)
        self.c = Communicate()

    @QtCore.pyqtSlot()
    def setModelData(self, editor, model, index):
        self.c.data_changed.emit(index, index)

最佳答案

As you can read hereQtCore.SIGNALPyQt4之后停产,因此不兼容。

This page解释了PyQt5的新型信号和插槽。语法为:

PyQt5.QtCore.pyqtSignal(types[, name[, revision=0[, arguments=[]]]])

您的案件可以翻译成:
from PyQt5 import pyqtsignal

data_changed = pyqtsignal(QModelindex,QModelIndex)

并发出您的信号:
self.data_changed.emit(index, index)

编辑:从下面的评论改编的解决方案。

09-25 20:10