詳解pyqt5 動(dòng)畫在QThread線程中無法運(yùn)行問題

更新時(shí)間：2018年05月05日 08:47:34 作者：蟬時(shí)雨

這篇文章主要介紹了詳解pyqt5 動(dòng)畫在QThread線程中無法運(yùn)行問題，小編覺得挺不錯(cuò)的，現(xiàn)在分享給大家，也給大家做個(gè)參考。一起跟隨小編過來看看吧

自己做了一個(gè)tcp工具，在學(xué)習(xí)動(dòng)畫的時(shí)候踩了坑，需求是根據(jù)上線變綠色，離線變灰色，如果連接斷開了，則變?yōu)榛疑?/p>

問題現(xiàn)象：

可以看到點(diǎn)擊“連接”，“離線”的時(shí)候動(dòng)畫是正常的，但是當(dāng)tcp超時(shí)斷開后，雖然離線按鈕變?yōu)檫B接了，卻沒有執(zhí)行離線動(dòng)畫

關(guān)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(jìn)制數(shù)據(jù)處理,兩個(gè)字節(jié)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(dòng)離線動(dòng)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(dòng)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(diǎn)：

經(jīng)過谷爹搜索，終于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(gè)安全的線程對(duì)象，我們不能直接在線程中去改變主程序的狀態(tài)，我們必須通過信號(hào)的方式去更新QGraphics

解決方法：

首先，我們編輯一個(gè)信號(hào)方法

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

然后添加相關(guān)信號(hào)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(gè)動(dòng)畫信號(hào)
      self.tcpth.start()

在線程中發(fā)出離線動(dòng)畫的信號(hào)

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(jìn)制數(shù)據(jù)處理,兩個(gè)字節(jié)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(gè)和QThread多線程收發(fā)消息原理一樣