這篇文章主要介紹pyqt5動(dòng)畫(huà)在QThread線程中無(wú)法運(yùn)行怎么辦，文中介紹的非常詳細(xì)，具有一定的參考價(jià)值，感興趣的小伙伴們一定要看完！
自己做了一個(gè)tcp工具，在學(xué)習(xí)動(dòng)畫(huà)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(kāi)了，則變?yōu)榛疑?/p>
問(wèn)題現(xiàn)象：
可以看到點(diǎn)擊“連接”，“離線”的時(shí)候動(dòng)畫(huà)是正常的，但是當(dāng)tcp超時(shí)斷開(kāi)后，雖然離線按鈕變?yōu)檫B接了，卻沒(méi)有執(zhí)行離線動(dòng)畫(huà)
關(guān)鍵源代碼如下
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重慶分公司
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(kāi)
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(huà)
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(wèn)題點(diǎn)：
經(jīng)過(guò)谷爹搜索，終于找到了問(wèn)題原因詳見(jiàn)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(guò)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(huà)信號(hào) self.tcpth.start()
在線程中發(fā)出離線動(dòng)畫(huà)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(kāi)
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
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