Pycall 抛出 ValidationError答案

【问题标题】：Pycall throws ValidationErrorPycall 抛出 ValidationError
【发布时间】：2019-01-12 10:52:39
【问题描述】：

我正在尝试在 atom 和 python3.6 中运行这些代码行：

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()

但我收到此错误：

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

如果您能帮我解决我的问题，我将不胜感激。提前谢谢你

【问题讨论】：

标签： python asterisk voip python-sip pycall

【解决方案1】：

查看source code for the validity check，似乎唯一可能引起您注意的检查是验证假脱机目录的检查。默认设置为/var/spool/asterisk/outgoing，但可以在创建调用文件时更改：

cf = CallFile(c, a, spool_dir='/my/asterisk/spool/outgoing')

【讨论】：