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SIPp / pysipp

SIPp for Humans - launch multiple agents with Python
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Error "Missing remote host parameter. This scenario requires it.\n' " #45

Closed access2praveen closed 5 years ago

access2praveen commented 5 years ago

I am encountering below error when i am trying to run sipp scenario file through pysipp

root@voiceqa-desktop:/opt/test# python3 sipp.py ('10.204.66.84', 5080) stderr for 'uac_AMR_BW' @ ('0.0.0.0', 34607) b'2018-10-08\t16:31:39.337407\t1538976699.337407: Missing remote host parameter. This scenario requires it.\n 'screen_file' contents for 'uac_AMR_BW' @ ('0.0.0.0', 34607): Traceback (most recent call last): File "sipp.py", line 8, in <module> scen() File "/usr/local/lib/python3.5/dist-packages/pysipp/agent.py", line 413, in __call__ raise_exc=raise_exc, **kwargs File "/usr/local/lib/python3.5/dist-packages/pluggy.py", line 724, in __call__ return self._hookexec(self, self._nonwrappers + self._wrappers, kwargs) File "/usr/local/lib/python3.5/dist-packages/pluggy.py", line 338, in _hookexec return self._inner_hookexec(hook, methods, kwargs) File "/usr/local/lib/python3.5/dist-packages/pluggy.py", line 333, in <lambda> _MultiCall(methods, kwargs, hook.spec_opts).execute() File "/usr/local/lib/python3.5/dist-packages/pluggy.py", line 596, in execute res = hook_impl.function(*args) File "/usr/local/lib/python3.5/dist-packages/pysipp/__init__.py", line 250, in pysipp_run_protocol finalize(cmds2procs, raise_exc=raise_exc) File "/usr/local/lib/python3.5/dist-packages/pysipp/__init__.py", line 228, in finalize raise SIPpFailure(msg) pysipp.SIPpFailure: Some agents failed 'uac_AMR_BW' with exit code 255 -> Command or syntax error: check stderr output

My Directory has only one scenario file

`root@voiceqa-desktop:/opt/test# ls /home/voiceqa/sipp-3.5.1/AMR_Testing/AMR-WB/ 

               uac_40sec_harvard.amr  uac_AMR_BW.xml  uac_Harvard_AMR.amr`

My python script looks as below

`import pysipp uas = pysipp.server(srcaddr=('10.204.66.84', 5080)) print(uas.srcaddr) uac = pysipp.client(destaddr=uas.srcaddr)

scen = pysipp.scenario()

scen = pysipp.scenario(dirpath='/home/voiceqa/sipp-3.5.1/AMR_Testing/AMR-WB',proxyaddr=('10.204.66.84', 5080))

print(scen.cmditems())

scen() pysipp.log_to_stderr('DEBUG') `

Kindly let me know how to fix this

Thanks Praveen

access2praveen commented 5 years ago

I have tried with latest version of pysipp as well , but it didnt resolve my problem , can you please help me how to iterate over scenario files , I have managed to run one instance of sipp with single scenario file with below code

`import pysipp

pysipp.log_to_stderr('DEBUG')

uac = pysipp.client(destaddr=('10.204.66.84',5080)) uac.local_host='10.204.66.30' uac.local_port='5030' uac.scen_file = '/home/voiceqa/sipp-3.5.1/AMR_Testing/AMR-WB/uac_AMR_BW.xml' uac()`

goodboy commented 5 years ago

I have tried with latest version of pysipp as well

@access2praveen can you try the branch from #46 though? Latest master doesn't have these changes merged in yet.

goodboy commented 5 years ago

ping @access2praveen did this end up solving your issue. Trying to get in fixes for an alpha release as per #42!

goodboy commented 5 years ago

Got no feedback on this so closing for now.