子对象通过信号/插槽从另一个线程发送

编程问答 2022-06-13

问题描述

我有一个带有QChartView的窗口，该图表上可以显示很多序列。如果我将所有系列添加到主线程上，那么它可能会阻塞一段时间，因此出于性能原因，我想在另一个线程中创建QChart，在其中添加所有系列并将QChart通过信号发送到显示窗口。>

简化示例：

# This Python file uses the following encoding: utf-8
import sys
from PyQt5.QtCore import pyqtSignal,pyqtSlot,QThread,Qt,QMetaObject
from PyQt5.QtWidgets import QApplication,QMainWindow
from PyQt5.QtChart import QChart,QChartView,QLineseries
import logging


class chart_window(QMainWindow):
    def __init__(self):
        QMainWindow.__init__(self)
        self.resize(640,480)
        chart = QChart()
        self.chartview = QChartView(self)
        self.chartview.setChart(chart)
        self.setCentralWidget(self.chartview)

    @pyqtSlot(QChart)
    def on_chart_created(self,chart):
        logging.debug("on_chart_created beg")
        logging.debug("self.thread= {0}".format(self.thread()))
        logging.debug("chartview.thread= {0}".format(self.chartview.thread()))
        logging.debug("current thread= {0}".format(QThread.currentThread()))
        logging.debug("chart.thread= {0}".format(chart.thread()))
        # self.chartview.setChart(chart)
        logging.debug("on_chart_created end")


class chart_creator(QThread):
    chart_created = pyqtSignal(QChart)

    @pyqtSlot()
    def delayed_start(self):
        self.start()

    def run(self):
        logging.debug("chart_creator.run beg")
        chart = QChart()
        series = QLineseries()
        series.setName("test")
        series.append(0,10)
        series.append(3,3)
        series.append(6,8)
        series.append(10,0)
        chart.addSeries(series)
        self.chart_created.emit(chart)
        logging.debug("chart_creator.run end")


if __name__ == "__main__":
    logging.basicConfig(format='%(asctime)s : %(levelname)s : %(message)s',level=logging.DEBUG)
    app = QApplication([])
    window = chart_window()
    creator = chart_creator()
    creator.chart_created.connect(window.on_chart_created)
    window.show()
    QMetaObject.invokeMethod(creator,"delayed_start",Qt.QueuedConnection)
    sys.exit(app.exec_())

输出如下：

2020-09-18 16:03:07,299 : DEBUG : chart_creator.run beg
2020-09-18 16:03:07,304 : DEBUG : chart_creator.run end
2020-09-18 16:03:07,304 : DEBUG : on_chart_created beg
2020-09-18 16:03:07,305 : DEBUG : self.thread= <PyQt5.QtCore.QThread object at 0x000001B9E62293A8>
2020-09-18 16:03:07,306 : DEBUG : chartview.thread= <PyQt5.QtCore.QThread object at 0x000001B9E62293A8>
2020-09-18 16:03:07,306 : DEBUG : current thread= <PyQt5.QtCore.QThread object at 0x000001B9E62293A8>
2020-09-18 16:03:07,307 : DEBUG : chart.thread= <__main__.chart_creator object at 0x000001B9E6226A68>
2020-09-18 16:03:07,307 : DEBUG : on_chart_created end

因此要明确一点：在'on_chart_created'中接收到的'chart'与主窗口及其所有子级（0x000001B9E62293A8）处于不同的线程（0x000001B9E6226A68）中。结果，如果我尝试做self.chartview.setChart(chart)，结果将是：

QObject: Cannot create children for a parent that is in a different thread.
(Parent is QTextDocument(0x1d45209f3b0),parent's thread is chart_creator(0x1d452022650),current thread is QThread(0x1d451bc14c0)

我尝试了movetoThread，但实际上没有用，而且我没有其他解决方案。

那么stackoverflow社区中有人知道如何解决此问题吗？

解决方法

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pyqt5pythonpython-multithreadingsignals-slots