pyqt5 動畫在QThread線程中無法運行問題

自己做了一個tcp工具，在學習動畫的時候踩了坑，需求是根據上線變綠色，離線變灰色，如果連接斷開了，則變為灰色

問題現象：

可以看到點擊“連接”，“離線”的時候動畫是正常的，但是當tcp超時斷開后，雖然離線按鈕變為連接了，卻沒有執行離線動畫

關鍵源代碼如下

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):
        """線程"""
        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:  # 十六進制數據處理,兩個字節隔開
                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("連接")
                self.scene1.offlineCol.start()  # 啟動離線動畫
                self.heartcheck.setChecked(False)
                self.heartcheck.setVisible(False)
                self.sendBtn.setDisabled(True)
            else:
                self.recv_signal.emit(tcpreceive)
            if stopsingle == 1:
                break

然后再啟動線程

            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()

 

問題點：

經過谷爹搜索，終於找到了問題原因詳見https://stackoverflow.com/questions/44328750/pyqt-qgraphicscene-move-item-in-background-thread

大致原因就是QGraphics Scene 不是一個安全的線程對象，我們不能直接在線程中去改變主程序的狀態，我們必須通過信號的方式去更新QGraphics

 

解決方法：

首先，我們編輯一個信號方法

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

然后添加相關信號槽

            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) # 添加一個動畫信號
            self.tcpth.start()

在線程中發出離線動畫的信號

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):
        """線程"""
        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:  # 十六進制數據處理,兩個字節隔開
                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("連接")
                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，這個和QThread多線程收發消息原理一樣