Question

我想将 QML activated中的ListView信号与 Python3 / PyQt5 pyqtSlot修饰方法相关联>（5.6）代码。

我目前的方法是通过QQmlApplicationEngine在我的代码中加载QML场景，然后使用findChild()获取对ListView的引用。

问题是，我只能在搜索像findChild(QObject, 'myList')这样的QObject时找到ListView。但是htis对象不允许我访问activated信号，很可能是因为此信号仅为QAbstractItemView及其后代定义。

因此，如果我尝试findChild(QListView, 'myList')，结果为None。因此，我无法获得activated信号。这是PyQt5中的一个错误还是我有另一种方式连接到这个信号？

这是一些最小的工作示例。

list.py：

import sys from OpenGL import GL from PyQt5.QtCore import QUrl, QObject from PyQt5.QtWidgets import QApplication, QListView from PyQt5.QtQml import QQmlApplicationEngine # Main Function if __name__ == '__main__': # Create main app app = QApplication(sys.argv) # Create QML engine engine = QQmlApplicationEngine(app) # Load the QML scene from file engine.load(QUrl('List.qml')) for root in engine.rootObjects(): node = root.findChild(QListView, 'myList') if node: # At this point I would like to connect something to the # node.activated signal print(node) # Execute the application and exit sys.exit(app.exec_())

List.qml：

import QtQuick 2.0 import QtQuick.Window 2.2 Window { visibility: Window.FullScreen visible: true ListView { objectName: "myList" anchors.fill: parent delegate: Item { width: parent.width * 0.8 height: 40 Row { id: row1 Rectangle { width: 40 height: 40 color: colorCode } Text { text: name font.bold: true anchors.verticalCenter: parent.verticalCenter } spacing: 10 } } model: ListModel { ListElement { name: "Grey" colorCode: "grey" } ListElement { name: "Red" colorCode: "red" } ListElement { name: "Blue" colorCode: "blue" } ListElement { name: "Green" colorCode: "green" } } } }

Answer 1

您可以使用QQuickView代替QQmlApplicationEngine来完成此操作。

我更改了你的python脚本，添加了一个继承自QQuickView的新类，并向名为“myList”的QML对象添加了一个信号。

此外，在QML中，我删除了Window类型的Item类型（您不能将Window与QQuickView一起使用）。如果要全屏显示应用程序，则必须将其指定为MyView类。在该示例中，如果单击其中一个彩色矩形，索引将显示在控制台中。

list.py：

import sys
from PyQt5.QtCore import QUrl, QObject
from PyQt5.QtWidgets import QApplication, QListView
from PyQt5.QtQuick import QQuickView, QQuickItem

class MyView(QQuickView):
    def __init__(self, parent=None):
        super().__init__(parent)
        # Load the QML scene from file
        self.setSource(QUrl('List.qml'))    
        #connect signal and source
        list = self.rootObject().findChild(QQuickItem, 'myList')
        list.mySignal.connect(self.mySlot)

    def mySlot(self, index):
        print(index)

# Main Function
if __name__ == '__main__':
    # Create main app
    app = QApplication(sys.argv)

    # Create QML view
    view = MyView()
    view.show()    

    # Execute the application and exit
    sys.exit(app.exec_())

List.qml：

import QtQuick 2.0
import QtQuick.Window 2.2

Item {
  width: 500
  height: 500
  ListView {
    anchors.fill: parent
    id: list
    objectName: "myList"
    signal mySignal(int index)
    delegate: Item {
      width: parent.width * 0.8
      height: 40
      Row {
        id: row1
        Rectangle {
          width: 40
          height: 40
          color: colorCode

          MouseArea{
            anchors.fill: parent
            onClicked: list.mySignal(index)
          }
        }

        Text {
          text: name
          font.bold: true
          anchors.verticalCenter: parent.verticalCenter
        }
        spacing: 10
      }
    }
    model: ListModel {
      ListElement {
        name: "Grey"
        colorCode: "grey"
      }

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

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

      ListElement {
        name: "Green"
        colorCode: "green"
      }
    }
  }

}

如何将PyQt5 pyqtSlot连接到QML ListView信号＆＃34;激活＆＃34;？

1 个答案: