稍后编辑：

Question

以下代码在self.emit行中断开。它在PyQt4中工作正常。如何修复此代码以便在PyQt5中工作？

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)

稍后编辑：

工作解决方案：

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)

Answer 1

As you can read here，QtCore.SIGNAL在PyQt4之后停止，因此不兼容。

This page解释了PyQt5的新式信号和广告位。语法是：

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

您的案件可以翻译为：

from PyQt5 import pyqtsignal

data_changed = pyqtsignal(QModelindex,QModelIndex)

并发出你的信号：

self.data_changed.emit(index, index)

编辑：根据以下评论改编的解决方案。

Answer 2

这在PyQt5中变得更加简单：

self.dataChanged.emit(index, index, [QtCore.Qt.EditRole])

如何在PyQt5中发出dataChanged

稍后编辑：

2 个答案: