pyqt5 动画在QThread线程中无法运行问题

自己做了一个tcp工具，在学习动画的时候踩了坑，需求是根据上线变绿色，离线变灰色，如果连接断开了，则变为灰色

问题现象：

可以看到点击“连接”，“离线”的时候动画是正常的，但是当tcp超时断开后，虽然离线按钮变为连接了，却没有执行离线动画

关键源代码如下

class BSJTcpThread(QtCore.QThread):
    recv_signal = QtCore.pyqtSignal(str)
    send_signal = QtCore.pyqtSignal(str)

    def __init__(self, socketcp, onBtn, heartcheck, senBtn, scene):
        super().__init__()
        self.s = socketcp
        self.yqtool = Bianlifunction()
        self.onBtn = onBtn
        self.heartcheck = heartcheck
        self.sendBtn = senBtn
        self.scene1 = scene

    def run(self):
        """线程"""
        global stopsingle
        stopsingle = 0
        while 1:
            btcpreceive = self.s.recv(1024)
            tcpreceive1 = str(binascii.b2a_hex(btcpreceive), encoding="utf-8")

            tcpreceive = ""
            i = 0
            while i < len(tcpreceive1) - 1:  # 十六进制数据处理,两个字节隔开
                if i == len(tcpreceive1) - 2:
                    tcpreceive += tcpreceive1[i:i + 2]
                    i += 2
                else:
                    tcpreceive += tcpreceive1[i:i + 2] + " "
                    i += 2

            if tcpreceive == "":
                stopsingle = 1
                self.s.shutdown(2)
                self.s.close()
                self.onBtn.setText("连接")
                self.scene1.offlineCol.start()  # 启动离线动画
                self.heartcheck.setChecked(False)
                self.heartcheck.setVisible(False)
                self.sendBtn.setDisabled(True)
            else:
                self.recv_signal.emit(tcpreceive)
            if stopsingle == 1:
                break

然后再启动线程

            self.tcpth = BSJTcpThread(self.s, self.onBtn, self.heartcheck, self.sendBtn, self.scene)
            self.tcpth.recv_signal.connect(self.fillrecvmsg)
            self.tcpth.send_signal.connect(self.fillsendmsg)
            self.tcpth.start()

 

问题点：

经过谷爹搜索，终于找到了问题原因详见https://stackoverflow.com/questions/44328750/pyqt-qgraphicscene-move-item-in-background-thread

大致原因就是QGraphics Scene 不是一个安全的线程对象，我们不能直接在线程中去改变主程序的状态，我们必须通过信号的方式去更新QGraphics

 

解决方法：

首先，我们编辑一个信号方法

    def threadAnimate(self, message):
        if message == "1":
            self.scene.offlineCol.start()

然后添加相关信号槽

            self.tcpth = BSJTcpThread(self.s, self.onBtn, self.heartcheck, self.sendBtn)
            self.tcpth.recv_signal.connect(self.fillrecvmsg)
            self.tcpth.send_signal.connect(self.fillsendmsg)
            self.tcpth.animate_signal.connect(self.threadAnimate)  # 添加一个动画信号
            self.tcpth.start()

在线程中发出离线动画的信号

class BSJTcpThread(QtCore.QThread):
    recv_signal = QtCore.pyqtSignal(str)
    send_signal = QtCore.pyqtSignal(str)
    animate_signal = QtCore.pyqtSignal(str)

    def __init__(self, socketcp, onBtn, heartcheck, senBtn):
        super().__init__()
        self.s = socketcp
        self.yqtool = Bianlifunction()
        self.onBtn = onBtn
        self.heartcheck = heartcheck
        self.sendBtn = senBtn

    def run(self):
        """线程"""
        global stopsingle
        stopsingle = 0
        while 1:
            btcpreceive = self.s.recv(1024)
            tcpreceive1 = str(binascii.b2a_hex(btcpreceive), encoding="utf-8")

            tcpreceive = ""
            i = 0
            while i < len(tcpreceive1) - 1:  # 十六进制数据处理,两个字节隔开
                if i == len(tcpreceive1) - 2:
                    tcpreceive += tcpreceive1[i:i + 2]
                    i += 2
                else:
                    tcpreceive += tcpreceive1[i:i + 2] + " "
                    i += 2

            if tcpreceive == "":
                stopsingle = 1
                self.s.shutdown(2)
                self.s.close()
                self.onBtn.setText("连接")
                self.animate_signal.emit("1")
                self.heartcheck.setChecked(False)
                self.heartcheck.setVisible(False)
                self.sendBtn.setDisabled(True)
            else:
                self.recv_signal.emit(tcpreceive)
            if stopsingle == 1:
                break

然后就可以了，这个和QThread多线程收发消息原理一样