From: News123 on

I captured a piece of code with a try except statement:

In the except part I display a stackdump

except Exception as e:
import traceback
ex_type,ex_value,e_b = sys.exc_info()
tbstring = traceback.format_exc()
print '%s%s:%s:%s' % \

The output, that I receive is:
File "C:\mycode\", line 63, in upload_chunk
rslt =
File "C:\Python26\lib\", line 1199, in __call__
return self.__send(self.__name, args)
File "C:\Python26\lib\", line 1489, in __request
File "C:\Python26\lib\", line 1253, in request
return self._parse_response(h.getfile(), sock)
File "C:\Python26\lib\", line 1387, in _parse_response
File "C:\Python26\lib\", 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\", 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.