在希望的田野上，我们看到了希望，看到了未来，看到了祖国未来繁荣昌盛的景象。
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
以上就是详解pyqt5 动画在QThread线程中无法运行问题。有志者，事竟成，破釜沉舟，百二秦关终属楚；苦心人，天不负，卧薪尝胆，三千越甲可吞吴！更多关于详解pyqt5 动画在QThread线程中无法运行问题请关注haodaima.com其它相关文章！
