将 QML 信号连接到 PySide2 插槽

Question

我有一些使用 Qt/C++ 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

我见过很多例子，但它们都不同，我猜 PyQt/PySide 现在变化很快

您能否提供将 QML 信号连接到 PySide Slot 的现代且简洁的方法？例如单击按钮以在 python 控制台中打印一些文本。这是我的简单代码示例

ma​​in.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_())

ma​​in.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
    }
}

Answer 1

在这些情况下的最佳实践是创建一个 QObject，将其导出到 QML 并在那里建立连接，就像在 C++ 中一样。

ma​​in.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_())

ma​​in.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 良好实践也适用于 Python/QML，只需极少的更改和限制。}

Answer 2

eyllanesc 的解决方案是直接同步调用方法test_slot。当test_slot 小而快时很好。但是如果包含很多操作，QML GUI 每次都会挂起，直到返回test_slot。

最类似于 Qt 的方式是 slog-signal 元对象连接（参见下面的 #CHANGES 和 //CHANGES）：

main.py

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

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

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

    app = QGuiApplication()
    foo = Foo()
    engine = QQmlApplicationEngine()
    
    #CHANGES: line excluded 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)
    
    #CHANGES: connect QML signal to Python slot
    engine.rootObjects()[0].test_signal.connect(foo.test_slot, type=Qt.ConnectionType.QueuedConnection)
    
    sys.exit(app.exec_())

main.qml

import QtQuick 2.13
import QtQuick.Controls 2.13

ApplicationWindow {
    visible: true
    
    //CHANGES: declare signal
    signal test_signal(string input_string)

    Button {
        anchors.centerIn: parent
        text: "Example"

        //CHANGES: emit signal
        onClicked: test_signal("Test string")
    }
}