XMPP连接到服务器(Python)
我目前正在学习一个教程/开始学习 我制作了EchoBot,但在运行时失败,错误如下:XMPP连接到服务器(Python),python,xmpp,bots,Python,Xmpp,Bots,我目前正在学习一个教程/开始学习 我制作了EchoBot,但在运行时失败,错误如下: File "echobot.py", line 59, in <module> if xmpp.connect(('talk.google.com'), '5222'): File "C:\Python31\Lib\sleekxmpp\clientxmpp.py", line 143, in connect reattempt=reattempt) File "C:\Python31\Lib\sle
File "echobot.py", line 59, in <module>
if xmpp.connect(('talk.google.com'), '5222'):
File "C:\Python31\Lib\sleekxmpp\clientxmpp.py", line 143, in connect
reattempt=reattempt)
File "C:\Python31\Lib\sleekxmpp\xmlstream\xmlstream.py", line 372, in connect
self.address = (host, int(port))
ValueError: invalid literal for int() with base 10: 'a'
谢谢你的提示/帮助 你写错了什么:
xmpp.connect(('talk.google.com', '5222'))
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there you missed a comma
第二件事是把港口放在第二位
xmpp.connect(('talk.google.com', 5222), '5222')
我不知道第二个论点,但是
互联网地址几乎总是
(主机、端口)
谢谢,是逗号!谢谢。这起作用了:xmpp.connect(('talk.google.com','5222'):
xmpp.connect(('talk.google.com', '5222'))
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there you missed a comma
xmpp.connect(('talk.google.com', 5222), '5222')