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Pycall抛出ValidationError
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Stack Overflow用户
提问于 2018-08-05 12:21:37
回答 1查看 301关注 0票数 2

我尝试在atompython3.6中运行以下代码行:

代码语言:javascript
复制
from pycall import CallFile, Call, Application
import sys


def call():
        c = Call('SIP/200')
        a = Application('Playback', 'hello-world')
        cf = CallFile(c, a)
        cf.spool()

if __name__ == '__main__':
        call()

但是我收到了这个错误:

代码语言:javascript
复制
Traceback (most recent call last):
  File "/home/pd/gits/voiphone/main.py", line 12, in <module>
    call()
  File "/home/pd/gits/voiphone/main.py", line 9, in call
    cf.spool()
  File "/home/pd/telephonerelayEnv/lib/python3.6/site-packages/pycall/callfile.py", line 135, in spool
    self.writefile()
  File "/home/pd/telephonerelayEnv/lib/python3.6/site-packages/pycall/callfile.py", line 123, in writefile
    f.write(self.contents)
  File "/home/pd/telephonerelayEnv/lib/python3.6/site-packages/pycall/callfile.py", line 118, in contents
    return '\n'.join(self.buildfile())
  File "/home/pd/telephonerelayEnv/lib/python3.6/site-packages/pycall/callfile.py", line 100, in buildfile
    raise ValidationError
pycall.errors.ValidationError

如果你能帮我解决我的问题,我将不胜感激。提前谢谢你

EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2018-08-14 02:05:07

查看source code for the validity check,似乎唯一能抓住您的检查就是验证假脱机目录的检查。默认情况下,它设置为/var/spool/asterisk/outgoing,但可以在创建调用文件时更改:

代码语言:javascript
复制
cf = CallFile(c, a, spool_dir='/my/asterisk/spool/outgoing')
票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/51691363

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