Locating the line number where an exception occurs in python code

Question

I have a code similar to this:

try:
  if x:
      statement1
      statement2
      statement3
  elif y:
      statement4
      statement5
      statement6
          


        

Answer 1

I believe the several answers here recommending you manage your try/except blocks more tightly are the answer you're looking for. That's a style thing, not a library thing.

However, at times we find ourselves in a situation where it's not a style thing, and you really do need the line number to do some other programattic action. If that's what you're asking, you should consider the traceback module. You can extract all the information you need about the most recent exception. The tb_lineno function will return the line number causing the exception.

>>> import traceback
>>> dir(traceback)
['__all__', '__builtins__', '__doc__', '__file__', '__name__', '__package__', '_format_final_exc_line', '_print', '_some_str', 'extract_stack', 'extract_tb', 'format_exc', 'format_exception', 'format_exception_only', 'format_list', 'format_stack', 'format_tb', 'linecache', 'print_exc', 'print_exception', 'print_last', 'print_list', 'print_stack', 'print_tb', 'sys', 'tb_lineno', 'types']
>>> help(traceback.tb_lineno)
Help on function tb_lineno in module traceback:

tb_lineno(tb)
Calculate correct line number of traceback given in tb.
Obsolete in 2.3

Newer versions of the traceback plumbing fix the issue prior to 2.3, allowing the code below to work as it was intended: (this is the "right way")

import traceback
import sys

try:
    raise Exception("foo")
except:
    for frame in traceback.extract_tb(sys.exc_info()[2]):
        fname,lineno,fn,text = frame
        print "Error in %s on line %d" % (fname, lineno)