如何获取发生python异常的实际行号？

Question

I have the following python decorator in my file decorators.py

def catch_exceptions(function):                                           #Line #1
    @wraps(function)                                                      #Line #2
    def decorator(*args, **kwargs):                                       #Line #3
        try:                                                              #Line #4
            return function(*args, **kwargs)                              #Line #5
        except Exception as e:                                            #Line #6
            exc_type, exc_obj, exc_tb = sys.exc_info()                    #Line #7
            fname = os.path.split(exc_tb.tb_frame.f_code.co_filename)[1]  #Line #8
            print "E=%s, F=%s, L=%s" % (str(e), fname, exc_tb.tb_lineno)  #Line #9
    return decorator                                                      #Line #10

In another file my_file.py , I use the catch_exceptions decorator like this:

from decorators import catch_exceptions                                   #Line #1
@catch_exceptions                                                         #Line #2
def my_method()                                                           #Line #3
    print (10/0 - 5/0)                                                    #Line #4

When I run it, I get the following output:

E=integer division or modulo by zero, F=decorators.py, L=5

Instead of it reporting the exception location as decorators.py , line #5, How can I get it to report the actual file and line number of where the exception originally occurred? That would be line #4 in my_file.py .

Answer 1

You may want to look into the traceback module

import traceback

try:
    function_that_raises_exception()
except Exception:
    traceback.print_exc()

It will print the entire stack trace.

Answer 2

If you want to do it as you described then

from functools import wraps
import sys, os, traceback

def catch_exceptions(function):                                           
    @wraps(function)                                                     
    def decorator(*args, **kwargs):                                      
        try:                                                              
            return function(*args, **kwargs)                              
        except Exception as e:                                            
            exc_type, exc_obj, exc_tb = sys.exc_info()
            print "E=%s, F=%s, L=%s" % (str(e), traceback.extract_tb(exc_tb)[-1][0], traceback.extract_tb(exc_tb)[-1][1]) ) 
    return decorator

But it's still traceback that you need to know.

I believe that filename that was printing was also a mistake.

So exc_tb is actual traceback object. And extracting it's data is made by extract_tb() it will do

Return a list of up to limit “pre-processed” stack trace entries extracted from the traceback object tb. It is useful for alternate formatting of stack traces. If limit is omitted or None , all entries are extracted. A “pre-processed” stack trace entry is a 4-tuple (filename, line number, function name*, text) representing the information that is usually printed for a stack trace.

So the second last element of traceback.extract_tb(exc_tb) would be exception that is raised in decorator, and the last would be in your function. So the last index( -1 ) is what we need. Then traceback.extract_tb(exc_tb)[-1][0] would be filename of (I suppose) your desired file, not decorators.py and traceback.extract_tb(exc_tb)[-1][1] would be the line when the exception was fired.

Answer 3

We can get the line number by splitting the string state of the traceback.format_exc() . Please have a look at my answer at the following question, I have added sample code there.

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