在Python中返回一个角色中的对象并在QML中获取另一个对象的引用

我正在写一个 Twitter 客户端。我实施了TweetItem and TweetModel。问题是有一个角色TweetItem called original。我希望它指向原始推文。

更新：我的代码中有一些拼写错误。现在我修复了它们。

import sys
from PyQt4 import QtCore, QtGui, QtDeclarative


class TweetModel(QtCore.QAbstractListModel):
    def __init__(self, prototype, parent=None):
        QtCore.QAbstractListModel.__init__(self, parent)
        self.setRoleNames(prototype.roleNames())
        self.tweets = []

    def appendRow(self, item):
        self.tweets.append(item)

    def rowCount(self, parent=QtCore.QModelIndex()):
        return len(self.tweets)

    def data(self, index, role):
        return self.tweets[index.row()].data(role)


class TweetItem(QtCore.QAbstractItemModel):
    def __init__(self, id=None, original=None, parent=None):
        QtCore.QAbstractItemModel.__init__(self, parent)
        self.idRole = QtCore.Qt.UserRole + 1
        # More Roles
        self.originalRole = QtCore.Qt.UserRole + 6

        self.id = id
        self.original = original

    def roleNames(self):
        names = {}
        names[self.idRole] = "id"
        names[self.originalRole] = "original"
        return names

    def data(self, role):
        if role == self.idRole:
            return self.id
        elif role == self.originalRole:
            # self.original == <__main__.TweetItem object at 0x7fb703d95d40>
            return self.original
        else:
            return None


if __name__ == "__main__":
    model = TweetModel(TweetItem())
    item = TweetItem("0001", None, model)
    model.appendRow(TweetItem("0002", item, model))

    App = QtGui.QApplication(sys.argv)
    view = QtDeclarative.QDeclarativeView()
    view.rootContext().setContextProperty("mymodel", model)
    view.setSource(QtCore.QUrl.fromLocalFile("main.qml"))
    view.show()
    App.exec_()

但我不能在 QML 中使用它。我得到一个undefined value.

import QtQuick 1.0

Rectangle {
  width: 360
  height: 360

  ListView {
         anchors.fill: parent
         model: mymodel
         // original.id == undefined
         delegate: Component { Text { text: id + " " + original.id } }
  } 
}

那么，是否可以在 a 中返回一个对象role并使用它？

正如所建议的dant3使用具有属性的 QObject。

以下是如何执行此操作的示例：

import sys
from PyQt4 import QtCore, QtGui, QtDeclarative
from PyQt4.QtCore import pyqtProperty, pyqtSignal, QObject


class TweetModel(QtCore.QAbstractListModel):
    def __init__(self, prototype, parent=None):
        QtCore.QAbstractListModel.__init__(self, parent)
        self.setRoleNames(prototype.roles)
        self.tweets = []

    def appendRow(self, item):
        self.tweets.append(item)

    def rowCount(self, parent=QtCore.QModelIndex()):
        return len(self.tweets)

    def data(self, index, role):
        return self.tweets[index.row()].data(role)


class TweetItem(QObject):
    roles = {
        QtCore.Qt.UserRole + 1: 'id',
        QtCore.Qt.UserRole + 6: 'original',
    }

    id_changed = pyqtSignal()

    def __init__(self, id=None, original=None, parent=None):
        QObject.__init__(self, parent=parent)

        self._data = {'original': original}
        self.id = id

    def data(self, key):
        return self._data[self.roles[key]]

    @pyqtProperty(str, notify=id_changed)
    def id(self):
        return self._data['id']

    @id.setter
    def id(self, value):
        if self._data.get('id') != value:
            self._data['id'] = value
            self.id_changed.emit()

if __name__ == "__main__":
    model = TweetModel(TweetItem)
    item = TweetItem("0001", None, model)
    model.appendRow(TweetItem("0002", item, model))

    App = QtGui.QApplication(sys.argv)
    view = QtDeclarative.QDeclarativeView()
    view.rootContext().setContextProperty("mymodel", model)
    view.setSource(QtCore.QUrl.fromLocalFile("main.qml"))
    view.show()
    App.exec_()

QML 文件保持不变。

我没有做original属性，因为您将其作为模型数据获取，但您可以以与id.