For some reason my code is having trouble opening a simple file:
This is the code:
file1 = open('recentlyUpdated.yaml')
And the error is:
IOError: [Errno 2] No such file or directory: 'recentlyUpdated.yaml'
open()
the full path to the file and none of it seems to work.The error "FileNotFoundError: [Errno 2] No such file or directory" is telling you that there is no file of that name in the working directory. So, try using the exact, or absolute path. In the above code, all of the information needed to locate the file is contained in the path string - absolute path.
The error FileNotFoundError Errno 2 no such file or directory occurs when Python cannot find the specified file in the current directory.
To create a file if not exist in Python, use the open() function. The open() is a built-in Python function that opens the file and returns it as a file object. The open() takes the file path and the mode as input and returns the file object as output.
log No such file or directory” the problem is most likely on the client side. In most cases, this simply indicates that the file or folder specified was a top-level item selected in the backup schedule and it did not exist at the time the backup ran.
os.listdir()
to see the list of files in the current working directoryos.getcwd()
(if you launch your code from an IDE, you may well be in a different directory)os.chdir(dir)
, dir
being the folder where the file is located, then open the file with just its name like you were doing.open
call.dir = r'C:\Python32'
'C:\\User\\Bob\\...'
'C:/Python32'
and do not need to be escaped.Let me clarify how Python finds files:
C:\Python\scripts
if you're on Windows.os.getcwd()
.If you try to do open('sortedLists.yaml')
, Python will see that you are passing it a relative path, so it will search for the file inside the current working directory.
Calling os.chdir()
will change the current working directory.
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
Most likely, the problem is that you're using a relative file path to open the file, but the current working directory isn't set to what you think it is.
It's a common misconception that relative paths are relative to the location of the python script, but this is untrue. Relative file paths are always relative to the current working directory, and the current working directory doesn't have to be the location of your python script.
You have three options:
Use an absolute path to open the file:
file = open(r'C:\path\to\your\file.yaml')
Generate the path to the file relative to your python script:
from pathlib import Path script_location = Path(__file__).absolute().parent file_location = script_location / 'file.yaml' file = file_location.open()
(See also: How do I get the path and name of the file that is currently executing?)
Change the current working directory before opening the file:
import os os.chdir(r'C:\path\to\your\file') file = open('file.yaml')
Other common mistakes that could cause a "file not found" error include:
Accidentally using escape sequences in a file path:
path = 'C:\Users\newton\file.yaml' # Incorrect! The '\n' in 'Users\newton' is a line break character!
To avoid making this mistake, remember to use raw string literals for file paths:
path = r'C:\Users\newton\file.yaml' # Correct!
(See also: Windows path in Python)
Forgetting that Windows doesn't display file extensions:
Since Windows doesn't display known file extensions, sometimes when you think your file is named file.yaml
, it's actually named file.yaml.yaml
. Double-check your file's extension.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With