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?
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。