Fileopen Error Python
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you have probably seen some. There are (at least) two distinguishable kinds of errors: syntax errors and exceptions. 8.1. Syntax Errors¶ Syntax errors, also known as parsing errors, are perhaps the most python exception message common kind of complaint you get while you are still learning Python: >>> while
Syntax For Generic Except Clause In Python
True print('Hello world') File "
Python Print Exception
and are not unconditionally fatal: you will soon learn how to handle them in Python programs. Most exceptions are not handled by programs, however, and result in error messages as shown here: >>> 10 * (1/0) Traceback (most recent call last): File "
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the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up Python's “open()” throws different errors https://docs.python.org/3/tutorial/errors.html for “file not found” - how to handle both exceptions? up vote 28 down vote favorite 6 I have a script where a user is prompted to type a filename (of a file that is to be opened), and if the file doesn't exist in the current directory, the user is prompted again. Here is the short version: file = input("Type filename: ") ... try: http://stackoverflow.com/questions/15032108/pythons-open-throws-different-errors-for-file-not-found-how-to-handle-b fileContent = open(filename, "r") ... except FileNotFoundError: ... When I tested my script on my MacOS X in Python 3.3x it worked perfectly fine when I type the wrong filename on purpose (it executes the suite under "expect"). However, when I wanted to run my code on a Windows computer in Python 3.2x, I get an error that says that "FileNotFoundError" is not defined. So, Python 3.2 on Windows thinks "FileNotFoundError" is a variable and the programs quits with an error. I figured out that Python 3.2 on Windows throws an "IOError" if the input filename is not valid. I tested it on my Linux machine in Python 2.7, and it's also an IOError. My problem is now, that the code with except "FileNotFoundError": won't run on Windows's Python 3.2, but if I change it to except "IOError": it won't work on my Mac anymore. How could I work around it? The only way I can think of is to use just except, which I usually don't want. python python-3.x filenotfoundexception ioerror share|improve this question edited Feb 22 '13 at 19:57 asked Feb 22 '13 at 19:48 user2015601 6 This isn't due to Mac/Wi
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more http://stackoverflow.com/questions/5627425/what-is-a-good-way-to-handle-exceptions-when-trying-to-read-a-file-in-python about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up What is a good way to handle exceptions when trying to read python exception a file in python? up vote 9 down vote favorite 2 I want to read a .csv file in python. I don't know if the file exists. My current solution is below. It feels sloppy to me because the two separate exception tests are awkwardly juxtaposed. Is there prettier way to do it? import csv fName = "aFile.csv" try: with open(fName, 'rb') as f: reader = csv.reader(f) for row fileopen error python in reader: pass #do stuff here except IOError: print "Could not read file:", fName python file-io exception-handling share|improve this question edited Apr 11 '11 at 21:08 asked Apr 11 '11 at 20:51 CharlesHolbrow 1,04931521 add a comment| 4 Answers 4 active oldest votes up vote 6 down vote accepted I guess I misunderstood what was being asked. Re-re-reading, it looks like Tim's answer is what you want. Let me just add this, however: if you want to catch an exception from open, then open has to be wrapped in a try. If the call to open is in the header of a with, then the with has to be in a try to catch the exception. There's no way around that. So the answer is either: "Tim's way" or "No, you're doing it correctly.". Previous unhelpful answer to which all the comments refer: import os if os.path.exists(fName): with open(fName, 'rb') as f: try: # do stuff except : # whatever reader errors you care about # handle error share|improve this answer edited Apr 11 '11 at 21:23 answered Apr 11 '11 at 20:55 Josh Caswell 52.4k11103152 5 Just because a file exists doesn't mean that you can read it! –Gabe Apr 11 '11 at