Correct Line number when doing exception handling in python

Question

I currently use this structure for capturing and recording my exception handling

1.  import sys
2.
3.  def break_python():
4.      test1 = 1/0
5.
6.  def main():
7.     #All main program functionality is included in this section
8.     break_python()
9.  
10. if __name__ == "__main__":
11.     try:
12.         main()
13.     except Exception as e:
14.         print dir(e)
15.         print e.message
16.         type1,details,tb = sys.exc_info_()
17.         print type1
18.         print details
19.         print tb
20.         line_no = tb.tb_lineno
21.         print "Line # " + str(line_no)
22.         print "Failed to run program"

When I run this as my primary program structure everything works however the line number will print as line 12 corresponding to the main() function call when I really want it to point to the function and line number of the individual error, in this case line 4 under function break_python()

Fabricator · Accepted Answer · 2014-06-02 18:43:18Z

2

I think you are looking for traceback.print_exc(file=sys.stdout) See https://docs.python.org/2/library/traceback.html#traceback-examples

answered Jun 2, 2014 at 18:43

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