There is already a much simpler way of doing this: > try: > i = int("string") > except ValueError as e: > print(e) > print("continued on") > j = int(9.0) > > The point of the 'try' block is to encapsulate the code you want to *stop* > executing if an exception is raised. If you want code to be run regardless > of whether an exception is raised, move it past the try-except. >
To be fair, I suspect the issue was there were two calls to int() there that might raise a ValueError, and the OP wanted to catch them with one except, so you would need to do somethign like: try: i = int("string") except ValueError as e: print(e) print("continued on") try: j = int(9.0) except ValueError as e: print(e) Which can seem a bit verbose, but in fact, there are a number of ways one might want to proceed with/without an error, and the current except, finally, else options cover them all in a clearly defined way. -CHB > > ~Amber > _______________________________________________ > Python-ideas mailing list > Python-ideas@python.org > https://mail.python.org/mailman/listinfo/python-ideas > Code of Conduct: http://python.org/psf/codeofconduct/ > -- Christopher Barker, Ph.D. Oceanographer Emergency Response Division NOAA/NOS/OR&R (206) 526-6959 voice 7600 Sand Point Way NE (206) 526-6329 fax Seattle, WA 98115 (206) 526-6317 main reception chris.bar...@noaa.gov
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