自己做了一個tcp工具,在學(xué)習(xí)動畫的時候踩了坑,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”的時候動畫是正常的,但是當(dāng)tcp超時斷開后,雖然離線按鈕變?yōu)檫B接了,卻沒有執(zhí)行離線動畫
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(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(tài),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(guān)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(fā)出離線動畫的信號
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