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I want to catch and log exceptions without exiting, e.g.,
try: do_stuff() except Exception as err: print(Exception, err) # I want to print the entire traceback here, # not just the exception name and details I want to print the exact same output that is printed when the exception is raised without the try..except intercepting the exception, and I do not want it to exit my program. How do I do this?
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To catch and print an exception that occurred in a code snippet, wrap it in an indented try block, followed by the command "except Exception as e" that catches the exception and saves its error message in string variable e . You can now print the error message with "print(e)" or use it for further processing.
Using printStackTrace() method − It print the name of the exception, description and complete stack trace including the line where exception occurred. Using toString() method − It prints the name and description of the exception. Using getMessage() method − Mostly used. It prints the description of the exception.
traceback.format_exc() or sys.exc_info() will yield more info if that's what you want.
import traceback import sys try: do_stuff() except Exception: print(traceback.format_exc()) # or print(sys.exc_info()[2]) If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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