I have a code similar to this:
try: if x: statement1 statement2 statement3 elif y: statement4 statement5 statement6 else: raise except: statement7 Here, I am sure that the exception occurs in If x: block, but I would like to know in which statement of If x: block the exception occurs. Is there a way to get the line number where the exception occurs?
Regards,
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 
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) 
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.
python -m pdb manage.py syncdb --settings=settings_development or python -m pdb manage.py runserver --settings=settings_development.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.
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!
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 >>> You should wrap the statements you care about more tightly. Extracting the line number from the traceback is going to be involved and fragile.
If you restructure the code like so, you should get a line number when the exception is raised again:
except: statement7 raise 
Using a general except statement is usually a bad programming practice, so you should specify in your except statement what exception you want to catch. ( like except ValueError: )
Moreover, you should surround with a try except structure the bits of code that are supposed to be raising an exception.
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.
we can get the line number by splitting the string state of the traceback.format_exc(). please try running the following code..
import traceback try: a = "str" b = 10 c = a + b except Exception as e: err_lineno = str(traceback.format_exc()).split(",")[1] print(err_lineno) this will produce the following output
line 7 