FileNotFoundError is a subclass of OSError, catch that or the exception itself:

except OSError as e:

Operating System exceptions have been reworked in Python 3.3; FileNotFoundError was added, and IOError has been merged into OSError. See the PEP 3151: Reworking the OS and IO exception hierarchy section in the What's New documentation.

For more details the OS Exceptions section for more information, scroll down for a class hierarchy.

That said, your code should still just work as IOError is now an alias for OSError:

>>> IOError
<class 'OSError'>

Make sure you are placing your exception handler in the correct location. Take a close look at the traceback for the exception to make sure you didn't miss where it is actually being raised. Last but not least, you did restart your Python script, right?

IOError is an alias for OSError. These used to be separate classes, but they were combined in Python 3.3 due to their confusing similarity.

FileNotFoundError is a subclass of OSError (aka IOError). This lets you do:

try:
    stream = open("non_existent_file", "rb")
    print("exists")
except FileNotFoundError:
    print("absent")
except IOError: # or equivalently, OSError
    print("exists, but can't be opened")

Note that you have to catch the more specific exception type first.

windows is going on.

backslashes are special characters in strings.

annoyingly windows uses them to seperate paths.

to print a backslash you need to escape it with a backslash

print("\\")

you may use:

import os
import json

filename = 'username.json'
username = input('What is your name ? ')

if os.path.isfile(filename ): # check if the file exist:
    if os.stat(filename ).st_size == 0: # check if the file is empty
        with open(filename, 'w') as fp:
            json.dump([username], fp)
            print('We will remember you as ' + username + ' !')

    else:
        with open(filename) as fp:
            user_names = json.load(fp)

        with open(filename, 'w') as fp:
            if username in user_names : 
                print('Welcome back ' + username + '!')
            else:
                user_names.append(username)
                json.dump(user_names, fp)
                print('We will remember you as ' + username + ' !')

else:
    with open(filename, 'w') as fp:
        json.dump([username])
        print('We will remember you as ' + username + ' !')

As the error message suggest, there is no file called "data.json" in the same directory as the one in which you are running the program. If you explicitly use the full path to the file, the program should work.

If FileNotFoundError isn't there, define it:

try:
    FileNotFoundError
except NameError:
    FileNotFoundError = IOError

Now you can catch FileNotFoundError in Python 2 since it's really IOError.

Be careful though, IOError has other meanings. In particular, any message should probably say "file could not be read" rather than "file not found."

The error clearly says that the file is not to be found. Try the following. 1. make sure that the filename 1.json is available from where you are calling the python interpretor.

for example: if you are calling $ python game/game.py, then the file should be in the present working directory, not in game dir

Try using absolute paths if you can

import os
base_dir = /path/to/json/dir
filename = str(id)+".json"
abs_file = os.path.join(base_dir, filename)
with open(abs_file, "r"):
    #do stuff

If you need the json files to be relative to the game.py file and still need the game file to be called from elsewhere, a good practice would be to define base_dir using __file__ attribute of the python file

base_dir = os.path.dirname(__file__)

The reason you're geting NoneType error is that somehow the loc variable is being set to None. which means that you are passing None to the Player's constructor. Since you haven't provided the code where you initialize player, I am assuming that you're passing the result of getRoom() as loc to the constructor. If that is the case, make sure that the value returned by getRoom is not None. you need an explicit return statement at the end of the function. return ret . by default any function without a return statement returns None. That could be your issue

it depends on the folder where you are running the script, not the script folder.

For get the real path to the directory where your script are, you can use it :

import os 
dir_path = os.path.dirname(os.path.realpath(__file__))

open(dir_path + '/' + 'data.json')

Pass in arguments:

import errno
import os

raise FileNotFoundError(
    errno.ENOENT, os.strerror(errno.ENOENT), filename)

FileNotFoundError is a subclass of OSError, which takes several arguments. The first is an error code from the errno module (file not found is always errno.ENOENT), the second the error message (use os.strerror() to obtain this), and pass in the filename as the 3rd.

The final string representation used in a traceback is built from those arguments:

>>> print(FileNotFoundError(errno.ENOENT, os.strerror(errno.ENOENT), 'foobar'))
[Errno 2] No such file or directory: 'foobar'

Let me clarify how Python finds files:

• An absolute path is a path that starts with your computer's root directory, for example C:\Python\scripts if you're on Windows.
• A relative path is a path that does not start with your computer's root directory, and is instead relative to something called the working directory.

If you try to do open('recentlyUpdated.yaml'), Python will see that you are passing it a relative path, so it will search for the file inside the current working directory.

To diagnose the problem:

• Ensure the file exists (and has the right file extension): use os.listdir() to see the list of files in the current working directory.
• Ensure you're in the expected directory using os.getcwd().
  (If you launch your code from an IDE, you may be in a different directory.)

You can then either:

• Call os.chdir(dir) where dir is the directory containing the file. This will change the current working directory. Then, open the file using just its name, e.g. open("file.txt").
• Specify an absolute path to the file in your open call.

By the way:

• Use a raw string (r"") if your path uses backslashes, like so: dir = r'C:\Python32'
  • If you don't use raw string, you have to escape every backslash: 'C:\\User\\Bob\\...'
  • Forward-slashes also work on Windows 'C:/Python32' and do not need to be escaped.

Example: Let's say file.txt is found in C:\Folder.

To open it, you can do:

os.chdir(r'C:\Folder')
open('file.txt')  # relative path, looks inside the current working directory

or

open(r'C:\Folder\file.txt')  # absolute path