将QML信号连接到PySide2插槽

小编典典

将QML信号连接到PySide2插槽

python

我对使用Qt / C ++有一定的经验，现在我想切换到PySide2 + QML。我想将ui信号（例如单击按钮）连接到python插槽

我已经看到了很多示例，但是它们都不同，我想PyQt / PySide现在正在迅速变化

您能为我提供一种将QML信号连接到PySide插槽的现代且干净的方法吗？例如，单击“按钮”以在python控制台中打印一些文本。这是我简单的代码示例

main.py

from PySide2.QtGui import QGuiApplication
from PySide2.QtQml import QQmlApplicationEngine

def test_slot(string): # pseudo slot
    print(string)

if __name__ == "__main__":
    app = QGuiApplication()
    engine = QQmlApplicationEngine('main.qml')
    exit(app.exec_())

main.qml

import QtQuick 2.13
import QtQuick.Controls 2.13

ApplicationWindow {
    visible: true

    Button {
        anchors.centerIn: parent
        text: "Example"
        onClicked: test_slot("Test") //pseudo signal
    }
}

阅读 138

2021-01-20

共1个答案

小编典典

在这些情况下，最佳实践是创建一个QObject，将其导出到QML并在那里进行连接，这也是在C ++中完成的。

main.py

from PySide2.QtCore import QObject, QUrl, Slot
from PySide2.QtGui import QGuiApplication
from PySide2.QtQml import QQmlApplicationEngine


class Foo(QObject):
    @Slot(str)
    def test_slot(self, string):
        print(string)


if __name__ == "__main__":
    import os
    import sys

    app = QGuiApplication()
    foo = Foo()
    engine = QQmlApplicationEngine()
    engine.rootContext().setContextProperty("foo", foo)
    qml_file = "main.qml"
    current_dir = os.path.dirname(os.path.realpath(__file__))
    filename = os.path.join(current_dir, qml_file)
    engine.load(QUrl.fromLocalFile(filename))
    if not engine.rootObjects():
        sys.exit(-1)
    sys.exit(app.exec_())

main.qml

import QtQuick 2.13
import QtQuick.Controls 2.13

ApplicationWindow {
    visible: true

    Button {
        anchors.centerIn: parent
        text: "Example"
        onClicked: foo.test_slot("Test")
    }
}

注意： 所有C ++ / QML良好实践也都适用于Python / QML，且更改和限制最少。

2021-01-20