使用Python从QML中动态创建的列表访问数据

时间:2019-03-29 19:53:57

标签: python qml pyqt5 qstringlistmodel

我在QML中有一个列表,并将其显示在listView对象中。按下按钮时,我需要从python访问此数据。在Python中,我创建了一个QStringListModel对象,并使用setContextProperty将其绑定到QML中的listModel。我可以看到正在QML中按预期方式创建和显示该列表,但是当我想从python访问数据时,该列表为空。这是代码:

QML:

import QtQuick 2.0
import QtQuick.Controls 2.3

Rectangle{
    id: root
    width:800
    height:600

    ListView {
        id: listView
        x: 476
        y: 64
        width: 110
        height: 160
        model: myModel
        ListModel {
            id: myModel
            ListElement {
                name: "Grey"
                colorCode: "grey"
            }

            ListElement {
                name: "Red"
                colorCode: "red"
            }

            ListElement {
                name: "Blue"
                colorCode: "blue"
            }

            ListElement {
                name: "Green"
                colorCode: "green"
            }
        }
        delegate: Item {
            x: 5
            width: 80
            height: 40
            Row {
                id: row1
                Rectangle {
                    width: 40
                    height: 40
                    color: colorCode
                }

                Text {
                    text: name
                    anchors.verticalCenter: parent.verticalCenter
                    font.bold: true
                }
                spacing: 10
            }
        }
    }
}

Python:

import sys
from PyQt5.QtCore import QUrl, QStringListModel
from PyQt5.QtWidgets import QApplication
from PyQt5.QtQuick import QQuickView


app = QApplication(sys.argv)
view = QQuickView()
view.setSource(QUrl("main.qml"))

pyList = QStringListModel()
view.rootContext().setContextProperty("myModel",pyList)
print(pyList.stringList())
print(pyList.rowCount())
view.show()
print("Done!")
sys.exit(app.exec_())

我给人的印象是,当我们使用python绑定时,在python中创建的对象将绑定到QML对象。因此,如果QML列表具有数据(在用户界面中动态创建),则python列表是否应自动填充该数据?我想念什么?

1 个答案:

答案 0 :(得分:1)

您假设由于通过setContextProperty()传递的模型与ListModel的名称相同而没有相同,因此,相反,由于QML重叠这两个名称,这会使您的程序不稳定。相反,您必须在python中创建一个模型并将其导出,但是由于您还想与QML进行交互,因此最好将具有模型的QObject导出为qproperty。不要使用findChild,findChildren在python中获取QML元素,而必须执行相反的操作,将元素从python导出到QML。

考虑到上述情况,除了可以从QML调用的插槽之外,我还实现了具有qproperty模型的Manager类。为了避免使代码复杂化,我使用QStandardItemModel类作为模型的基础,QStringListModel是只读模型,因此在这种情况下不起作用。

main.py

from enum import Enum
from PyQt5 import QtCore, QtGui, QtQuick

class ElementRoles:
    NameRole = QtCore.Qt.UserRole + 1000
    ColorRole = QtCore.Qt.UserRole + 1001

class ElementModel(QtGui.QStandardItemModel, ElementRoles):
    QtCore.Q_ENUM(ElementRoles)

    def __init__(self, parent=None):
        super(ElementModel, self).__init__(parent)
        roles = {
            ElementModel.NameRole: b'name',
            ElementModel.ColorRole: b'colorCode'
        }
        self.setItemRoleNames(roles)

    @QtCore.pyqtSlot(str, QtGui.QColor)
    def addElement(self, name, color):
        item = QtGui.QStandardItem()
        item.setData(name, ElementModel.NameRole)
        item.setData(color, ElementModel.ColorRole)
        self.appendRow(item)

class Manager(QtCore.QObject):
    def __init__(self, parent=None):
        super(Manager, self).__init__(parent)
        self._model = ElementModel()

    @QtCore.pyqtProperty(QtCore.QObject, constant=True)
    def model(self):
        return self._model

    @QtCore.pyqtSlot()
    def on_clicked(self):
        print("count:", self._model.rowCount())
        for row in range(self._model.rowCount()):
            it = self._model.item(row)
            print("row:", row)
            for role, name in self._model.roleNames().items():
                print("role:", name, "data:", it.data(role))

if __name__ == '__main__':
    import os
    import sys
    app = QtGui.QGuiApplication(sys.argv)
    manager = Manager()
    view = QtQuick.QQuickView()
    file = os.path.join(os.path.dirname(os.path.realpath(__file__)), "main.qml")
    view.rootContext().setContextProperty("manager", manager)
    view.setSource(QtCore.QUrl.fromLocalFile(file))
    view.show()
    sys.exit(app.exec_())

main.qml

import QtQuick 2.0
import QtQuick.Controls 2.3
import QtQuick.Layouts 1.3

Rectangle{
    id: root
    width:800
    height:600

    ColumnLayout{
        anchors.fill: parent
        ListView {
            id: listView
            Layout.alignment: Qt.AlignLeft
            Layout.fillHeight: true
            model: manager.model
            delegate: Item {
                x: 5
                width: 80
                height: 40
                Row {
                    id: row1
                    Rectangle {
                        width: 40
                        height: 40
                        color: colorCode
                    }

                    Text {
                        text: name
                        anchors.verticalCenter: parent.verticalCenter
                        font.bold: true
                    }
                    spacing: 10
                }
            }
        }

        Button{
            Layout.alignment: Qt.AlignCenter
            text: "click me"
            onClicked: manager.on_clicked()
        }
    }

    Component.onCompleted:{
        manager.model.addElement("Gray", "gray")
        manager.model.addElement("Red", "red")
        manager.model.addElement("Blue", "blue")
        manager.model.addElement("Green", "green")
    }
}

我建议您阅读my another answer以获得更详细的说明。

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