宜配屋

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