News123
3/25/2010 8:25:00 AM
Hi,
I captured a piece of code with a try except statement:
In the except part I display a stackdump
try:
domyxmlrpcstuff()
except Exception as e:
import traceback
ex_type,ex_value,e_b = sys.exc_info()
tbstring = traceback.format_exc()
print '%s%s:%s:%s' % \
(msg,ex_type,ex_value,tbstring)
The output, that I receive is:
File "C:\mycode\myrpcclient.py", line 63, in upload_chunk
rslt = myrpcclient.call()
File "C:\Python26\lib\xmlrpclib.py", line 1199, in __call__
return self.__send(self.__name, args)
File "C:\Python26\lib\xmlrpclib.py", line 1489, in __request
verbose=self.__verbose
File "C:\Python26\lib\xmlrpclib.py", line 1253, in request
return self._parse_response(h.getfile(), sock)
File "C:\Python26\lib\xmlrpclib.py", line 1387, in _parse_response
p.feed(response)
File "C:\Python26\lib\xmlrpclib.py", line 601, in feed
self._parser.Parse(data, 0)
ExpatError: syntax error: line 1, column 0
In order to understand more I would like to display the value of
data in C:\Python26\lib\xmlrpclib.py", line 601
Is this possible in a non interactive fashion?
This is a generic question about inspecting variables down the stack,
whenever an exception occurs.
I started another thread specifically about displaying the invalid
xmlrpc data.
N