return python dict to QML (PySide2)

Question

我正在尝试找到一种从PySide2.QtCore.Slot返回python字典的方法。

main.py

import sys
from PySide2.QtCore import QObject, Slot
from PySide2.QtGui import QGuiApplication, QQmlApplicationEngine
class Backend(QObject):  
def __init__(self, parent=None):
return super().__init(parent)
@Slot(result=QObject)
def get_data(self):
data = {}
data["info1"] = "some information"
data["info2"] = "some  more information"
data["info3"] = 42
return data
if __name__ == '__main':
BACKEND = Backend()
APP = QGuiApplication(sys.argv)
ENGINE = QQmlApplicationEngine(APP)
ENGINE.rootContext().setContextProperty('backend', BACKEND)
ENGINE.load("main.qml")
sys.exit(APP.exec_())

main.qml：

import QtQuick 2.4
import QtQuick.Controls 1.4
ApplicationWindow {
id: root
width: 640
height: 480
visible: true
color: "#F0F0F0"
title: qsTr("Test")
Text {
anchors.centerIn: parent
text: backend.get_data()["info1"]
}
} 

我认为它是以某种方式在 QAbstractItemModel.roleNames(( 中完成的，因为它返回一个QHash<int, QByteArray>？

如果它不能像这样工作，任何人都可以支持我在python后端和QML前端之间交换信息的"正确方法"吗？

提前致谢:)

Answer 1

python 的基本类型在导出到 QML 时会转换为相应的类型，因为它们是受支持的，但是要使Slot()返回某些内容，数据类型必须通过result参数指示，在此QVariant中作为字符串。

例：

main.py

from PySide2 import QtCore, QtGui, QtQml

class Helper(QtCore.QObject):
@QtCore.Slot(result='QVariant')
def foo(self):
return {"a": 1, "b": 2}

if __name__ == '__main__':
import sys
app = QtGui.QGuiApplication(sys.argv)
engine = QtQml.QQmlApplicationEngine()
helper = Helper()
engine.rootContext().setContextProperty("helper", helper)
engine.load(QtCore.QUrl.fromLocalFile('main.qml'))
if not engine.rootObjects():
sys.exit(-1)
sys.exit(app.exec_())

主.qml

import QtQuick 2.9
import QtQuick.Controls 2.4
ApplicationWindow {
visible: true
Component.onCompleted: { 
var data = helper.foo()
for(var key in data){
var value = data[key]
console.log(key, ": ", value)
}
}
}

输出：

qml: a :  1
qml: b :  2