When i want to open a file in C:\ (windows dir.) this error shows up:
PermissionError: [Errno 13] Permission denied: 'C:\h.txt'
What should i do?
I know this question has been asked several times but i can't find solution!
code:
f=open ('C:\\h.txt','w')
f.write ('python')
f.close
I am not on win machine, but give this a try, you can manage permissions using these commands
Try to open your file using os.fdopen
import os
with os.fdopen(os.open('file.txt', os.O_WRONLY | os.O_CREAT, 0600), 'w') as f:
f.write(...)
UPDATE
import os
is_accessible = os.access("C:\\temp\\python",os.F_OK) #Check if you have access, this should be a path
if is_accessible == False: #If you don't, create the path
os.makedirs("C:\\temp\\python")
os.chdir("C:\\temp\\python") # Check now if the path exist
f = os.open( "p.txt", os.O_RDWR|os.O_CREAT ) #Create the file
os.write(f, b"This is a test \n") #Try to write
os.close(f)
I'm not on a Windows machine but perhaps you should try and create this file in the directory c:\Temp.
Likewise make sure you've not got Notepad etc with that file open.
Related
studying foundation year ComSci and I'm trying to get a file to open. Here's my code:
def main():
filename = raw_input("Please enter file name ")
infile = open(filename,'r')
data = infile.read()
print data
main()
I believe the code is correct but when I try and open a file, for example
C:\Users\Manol\OneDrive\Documents\Uni\Programming\File Processing File
It comes back with
Traceback (most recent call last):
File "C:\Users\Manol\OneDrive\Documents\Uni\Programming\File Processing File\FirstFileProcessingRead.py", line 6, in <module>
main()
File "C:\Users\Manol\OneDrive\Documents\Uni\Programming\File Processing File\FirstFileProcessingRead.py", line 3, in main
infile = open(filename,'r')
IOError: [Errno 13] Permission denied: 'C:\\Users\\Manol\\OneDrive\\Documents\\Uni\\Programming\\File Processing File'
Aside from your permissions problem, does open() not need the file extension to function correctly?
aa = open("C:\\aaa\\bbb\\ccc\\ddd.json","r")
will work, whereas
aa = open("C:\\aaa\\bbb\\ccc\\ddd","r")
will return
FileNotFoundError: [Errno 2] No such file or directory: 'C:\\aaa\\bbb\\ccc\\ddd'
As a way of testing folder permissions - copy the file to the same area as your .py and try opening it there. If you've still got same problems, then does your account have rights to the file? (right-click >> security)
See if file is encrypted and put two Backslash ( \\ ) or replace to Forward slash ( / ) and put extension of file
I change a few your code.
def main():
print("Please enter file name:")
filename = input()
infile = open(filename,'r')
data = infile.read()
print(data)
main()
Problem:
While installation when you select install python only for me then this issue is likely to happen because python is not granted with the permission to read every location, in your case One Drive.
Solution:
Place the file in your python file's root folder i.e inside the folder where you have placed the python file, then this will work.
You are also not providing extension (.txt). Don't forget to provide it while inputting the data.
I am trying to read content of a file on my work network from my work network. I copy and pasted a code snippet from a google search and modified it to the below. Why might I still be getting [Errno 2] (I have changed some of the path names for this question board)
The file path in my file explorer shows that "> This PC > word common common" and I don't have "This PC" in my path. I tried adding that into the place I would think it goes in the string. That didn't solve it.
I tried making sure I have matching capitalization. That didn't solve it.
I tried renaming the file to have a *.txt on the end. That didn't solve it.
I tried the different variations of // and / and \ with and without the r predecessor and while that did eliminate the first error I was getting. It didn't help this error.
(Looking at the code errors in the right gutter is says my line length is greater than the PEP8 standard. While I doubt that is the root of my problem, if you can throw in the 'right' wrap method for a file path that long that would be helpful.)
myfile = open("z:/abcdefg/abc123_proj2/word_general/word common common/Users/Mariee/Python/abc_abc_ab_Full_Report_12345-1_R9999_962019_9246", "rt") # open lorem.txt for reading text
contents = myfile.read() # read the entire file into a string
myfile.close() # close the file
print(contents) # print contents
Full Error Copy:
C:\Users\e087680\PycharmProjects\FailureCompiling\venv\Scripts\python.exe C:/Users/e087680/PycharmProjects/FailureCompiling/FirstScriptAttempt.py
Traceback (most recent call last):
File "C:/Users/e087680/PycharmProjects/FailureCompiling/FirstScriptAttempt.py", line 1, in
myfile = open("z:/abcdefg/abc123_proj2/word_general/word common common/Users/Mariee/Python/abc_abc_ab_Full_Report_12345-1_R9999_962019_9246", "rt") # open lorem.txt for reading text
FileNotFoundError: [Errno 2] No such file or directory: 'z:/abcdefg/abc123_proj2/word_general/word common common/Users/Mariee/Python/abc_abc_ab_Full_Report_12345-1_R9999_962019_9246'
EDIT
DEBUG EFFORTS
working to figure out how to change directory. Just in case that is the problem. Tested this code bit
import os
path = "z:/abcdefg/abc123_proj2/word_general/word common common/Users/Mariee/Python/abc_abc_ab_Full_Report_12345-1_R9999_962019_9246"
os.chdir(path)
isExist = os.path.exists(path)
print(isExist)
Received this error
C:\Users\e087680\PycharmProjects\FailureCompiling\venv\Scripts\python.exe C:/Users/e087680/PycharmProjects/FailureCompiling/ScriptDebugJunkFile.py
Traceback (most recent call last):
File "C:/Users/e087680/PycharmProjects/FailureCompiling/ScriptDebugJunkFile.py", line 5, in <module>
os.chdir(path)
FileNotFoundError: [WinError 3] The system cannot find the path specified: 'z:/abcdefg/abc123_proj2/word_general/word common common/Users/Mariee/Python/abc_abc_ab_Full_Report_12345-1_R9999_962019_9246'
My intention for adding the picture below is to show how File Explorer displays the file path for my file
FileExplorerPathScreenShot
EDIT
I think this confirms that my 'OS' doesn't have my file.
from os import path
path.exists("PCC_ESS_FC_Full_Report_65000122-1_R0016_962019_9246")
def main():
print ("File exists:" + str(path.exists('PCC_ESS_FC_Full_Report_65000122-1_R0016_962019_9246')))
if __name__== "__main__":
main()
Output
File exists: False
I thought OS was a standard variable for Operating system. Now I'm not sure.
EDIT
Using Cmd in DOS, I confirmed that my path for the z: is correct
EDIT - Success
I ran
import os
print( os.listdir("z:/"))
Confirmed I don't need the monster string of folders.
Confirmed, although explorer doesn't show it, it is a *.txt file
Once I implemented these two items the first code worked fine.
Thank you #Furas
To open and read a file specify the filename in your path:
myfile = open("U:/matrix_neo/word common common/hello world.txt", "rt") # open file
contents = myfile.read() # read the entire file into a string
myfile.close() # close the file
print(contents) # print contents
The U: is a mapped drive in my network.
I did not find any issue with your change dir example. I used a path on my U: path again and it returned True.
import os
path = "U:/matrix_neo/word common common"
os.chdir(path)
isExist = os.path.exists(path)
print(isExist)
The check the attributes on the directory that you are trying to read from. Also try to copy the file to a local drive for a test and see if you can read the file and also check if it exists.
This is an alternative to the above and uses your path to make sure that the long file path works:
import os
mypath = "z:/abcdefg/abc123_proj2/word_general/word common common/Users/Mariee/Python/abc_abc_ab_Full_Report_12345-1_R9999_962019_9246"
myfile = 'whatever is your filename.txt'
if not os.path.isdir(mypath):
os.makedirs (mypath)
file_path = os.path.join(mypath, myfile)
print(file_path)
if os.path.exists(file_path) is True:
with open(file_path) as filein:
contents = filein.read()
print(contents)
I tested this code using a long csv file.,Replace the variable myfile with whatever is your file name.
I'm having an issue trying to open a file that is definitely saved to my computer ('NYT-bestsellers.txt'), but whenever I try opening it with my code I get the error
FileNotFoundError: [Errno 2] No such file or directory: 'NYT-bestsellers.txt'
I thought about using the method where you use the full path to open the file… but this is part of an assignment that I'll be submitting later this week. If I open the file using a specific path from my laptop, I'm worried that it won't open for the marker. Please advise!
with open('NYT-bestsellers.txt', 'r') as file:
file = file.splitlines()
As Ryan said, every time you open a file by a relative name, you need to make clear for the current work path.
import sys
import os
current_work_directory = os.getcwd() # Return a string representing the current working directory.
print('Current work directory: {}'.format(current_work_directory))
# Make sure it's an absolute path.
abs_work_directory = os.path.abspath(current_work_directory)
print('Current work directory (full path): {}'.format(abs_work_directory))
print()
filename = 'NYT-bestsellers.txt'
# Check whether file exists.
if not os.path.isfile(filename):
# Stop with leaving a note to the user.
print('It seems file "{}" not exists in directory: "{}"'.format(filename, current_work_directory))
sys.exit(1)
# File exists, go on!
with open(filename, 'r') as file:
file = file.splitlines()
If you confirm that the file will be along with your python script file, you can do some preparatory work before opening the file:
script_directory = os.path.split(os.path.abspath(__file__))[0]
print(script_directory)
abs_filename = os.path.join(script_directory, filename)
print(abs_filename)
with open(abs_filename, 'r') as file:
file = file.splitlines()
im have FLASK app with
www/FlaskApp/FlaskApp/init.py file with funtion
python file wut next contains
#app.route('/')
def hello():
file = open('myfile.txt', 'w+')
os.mknod("newfile.txt")
return render_template('page2.html')
but if im run site,its return error, in file log write
PermissionError: [Errno 13] Permission denied: 'myfile.txt'
im set permision 777 for all www directories
open FileZilla
right click on www dir, and set 777 permision
Why file dont create?
Not sure if this is an optimal solution, and I don't know enough about Flask as to tell you why the relative path isn't working (I would think that it would write the file where ever your python script was) but you could get it to work by using an environment variable to specify where to store your apps data. For instance:
import os
#app.route('/')
def open_file():
filename = os.path.join(os.environ['FLASK_APP_DATA'], 'myfile.txt')
print (filename)
file = open(filename, 'w+')
file.write("This is a test")
file.close()
Then you could have the environment variable set differently on your dev box and your prod box.
I am attempting to create and write to a temporary file on Windows OS using Python. I have used the Python module tempfile to create a temporary file.
But when I go to write that temporary file I get an error Permission Denied. Am I not allowed to write to temporary files?! Am I doing something wrong? If I want to create and write to a temporary file how should should I do it in Python? I want to create a temporary file in the temp directory for security purposes and not locally (in the dir the .exe is executing).
IOError: [Errno 13] Permission denied: 'c:\\users\\blah~1\\appdata\\local\\temp\\tmpiwz8qw'
temp = tempfile.NamedTemporaryFile().name
f = open(temp, 'w') # error occurs on this line
NamedTemporaryFile actually creates and opens the file for you, there's no need for you to open it again for writing.
In fact, the Python docs state:
Whether the name can be used to open the file a second time, while the named temporary file is still open, varies across platforms (it can be so used on Unix; it cannot on Windows NT or later).
That's why you're getting your permission error. What you're probably after is something like:
f = tempfile.NamedTemporaryFile(mode='w') # open file
temp = f.name # get name (if needed)
Use the delete parameter as below:
tmpf = NamedTemporaryFile(delete=False)
But then you need to manually delete the temporary file once you are done with it.
tmpf.close()
os.unlink(tmpf.name)
Reference for bug: https://github.com/bravoserver/bravo/issues/111
regards,
Vidyesh
Consider using os.path.join(tempfile.gettempdir(), os.urandom(24).hex()) instead. It's reliable, cross-platform, and the only caveat is that it doesn't work on FAT partitions.
NamedTemporaryFile has a number of issues, not the least of which is that it can fail to create files because of a permission error, fail to detect the permission error, and then loop millions of times, hanging your program and your filesystem.
The following custom implementation of named temporary file is expanded on the original answer by Erik Aronesty:
import os
import tempfile
class CustomNamedTemporaryFile:
"""
This custom implementation is needed because of the following limitation of tempfile.NamedTemporaryFile:
> Whether the name can be used to open the file a second time, while the named temporary file is still open,
> varies across platforms (it can be so used on Unix; it cannot on Windows NT or later).
"""
def __init__(self, mode='wb', delete=True):
self._mode = mode
self._delete = delete
def __enter__(self):
# Generate a random temporary file name
file_name = os.path.join(tempfile.gettempdir(), os.urandom(24).hex())
# Ensure the file is created
open(file_name, "x").close()
# Open the file in the given mode
self._tempFile = open(file_name, self._mode)
return self._tempFile
def __exit__(self, exc_type, exc_val, exc_tb):
self._tempFile.close()
if self._delete:
os.remove(self._tempFile.name)
This issue might be more complex than many of you think. Anyway this was my solution:
Make use of atexit module
def delete_files(files):
for file in files:
file.close()
os.unlink(file.name)
Make NamedTemporaryFile delete=False
temp_files = []
result_file = NamedTemporaryFile(dir=tmp_path(), suffix=".xlsx", delete=False)
self.temp_files.append(result_file)
Register delete_files as a clean up function
atexit.register(delete_files, temp_files)
tempfile.NamedTemporaryFile() :
It creates and opens a temporary file for you.
f = open(temp, 'w') :
You are again going to open the file which is already open and that's why you are getting Permission Denied error.
If you really wants to open the file again then you first need to close it which will look something like this-
temp= tempfile.NamedTemporaryFile()
temp.close()
f = open(temp.name, 'w')
Permission was denied because the file is Open during line 2 of your code.
close it with f.close() first then you can start writing on your tempfile