getting source code line of error?

[Paolo G. Cantore]

Am 20.11.21 um 20:15 schrieb Ulli Horlacher:
> Stefan Ram <ram at zedat.fu-berlin.de> wrote:
>> ram at zedat.fu-berlin.de (Stefan Ram) writes:
>>> except Exception as inst:
>>>     print( traceback.format_exc() )
>>
>>    More to the point of getting the line number:
> 
> As I wrote in my initial posting:
> I already have the line number. I am looking for the source code line!
> 
> So far I use:
> 
>        m = re.search(r'\n\s*(.+)\n.*\n$',traceback.format_exc())
>        if m: print('%s %s' % (prefix,m.group(1)))
> 
Stefan Ram's solution missed only the line content. Here it is.


import sys
import traceback

try:
     1/0
except ZeroDivisionError as exception:
     tr = traceback.TracebackException.from_exception( exception )
     x = tr.stack[0]
     print("Exception %s in line %s: %s" % (exception, x.lineno, x.line))


The traceback object does not only contain the lineno but also the 
content of the offending line.

--------------
Paolo