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've done the following before:

try:
    doing = "statement1"
    statement1
    doing = "statement2"
    statement2
    doing = "statement3"
    statement3
    doing = "statement4"
    statement4

 except:
    print "exception occurred doing ", doing

The advantage over printing checkpoints is there's no log output unless there actually is an exception.

Answer 2

You should run your program in a debugger, such as pdb. This will allow you to run your code normally, and then examine the environment when something unexpected like this occurs.

Given a script named 'main.py', run it like this:

python -m pdb main.py

Then, when your program starts, it will start in the debugger. Type c to continue until the next breakpoint (or crash). Then, you can examine the environment by doing things like print spam.eggs. You can also set breakpoints by doing pdb.set_trace() (I commonly do import pdb; pdb.set_trace()).

Additionally, what do you mean that it is "okay" for 'statement 3' to raise the exception? Are you expecting the exception? If so, it might be better to write a try/except block around this statement, so that the program can continue.

Answer 3

Building on JJ above..

The advantage of using system errors over statements is they record more specific information which will aid debugging later (believe me I get a lot)

eg. I record them to a text file, so after my programs have automatically run overnight on the server, I can retrieve any issues, and have enough information to quicken the repair!

More Info... Traceback & Sys

import traceback
import sys

try:
    print 1/0

except Exception as e:
    print '1', e.__doc__
    print '2', sys.exc_info()
    print '3', sys.exc_info()[0]
    print '4', sys.exc_info()[1]
    print '5', sys.exc_info()[2], 'Sorry I mean line...',traceback.tb_lineno(sys.exc_info()[2])
    ex_type, ex, tb = sys.exc_info()
    print '6', traceback.print_tb(tb)

Yields

>  1  Second argument to a division or modulo operation was zero. 
>  2  (<type 'exceptions.ZeroDivisionError'>, ZeroDivisionError('integer division
>      or modulo by zero',), <traceback object at 0x022DCF30>) 
>  3  <type 'exceptions.ZeroDivisionError'> 
>  4  integer division or modulo by zero 
>  5  <traceback object at 0x022DCF30> Sorry I mean line... 5
>  6  File "Z:\Programming\Python 2.7\Error.py", line 5, in <module>
>     print 1/0 
      None
>>>

Answer 4

You should wrap the statements you care about more tightly. Extracting the line number from the traceback is going to be involved and fragile.

Answer 5

Edit your source code, so that you remove one line at a time, until the error disappears, and that should point you closer to the problem.

Answer 6

what about this:

try:
  if x:
      print 'before statement 1'
      statement1
      print 'before statement 2' #ecc. ecc.
      statement2
      statement3
  elif y:
      statement4
      statement5
      statement6
  else:
      raise

except:
      statement7

this is the straightforward workaround but I suggest to use a debugger

or even better, use the sys module :D

try:
      if x:
          print 'before statement 1'
          statement1
          print 'before statement 2' #ecc. ecc.
          statement2
          statement3
      elif y:
          statement4
          statement5
          statement6
      else:
          raise
except:
    print sys.exc_traceback.tb_lineno 
    #this is the line number, but there are also other infos