Im creating a new thread to ask this question because i have tried every solution i found but non seems to be working for me. I want to output pdf files stored in media sub directory. #Compadre suggested some links that really helped me understand how the file system works in python but so solutions provided in the threads are not working. i want to output files in the following directory structure:
media/QuestionPapers/filename.pdf
view function for outputting the file is:
def Display(request, file_name):
File_Name = file_name.replace('_', ' ')
file_path = os.path.join(settings.MEDIA_ROOT, 'QuestionPapers',File_Name)
with open(file_path,'rb') as pdf:
response = HttpResponse(pdf.read(), content_type = 'application/pdf')
response['Content-Disposition'] = 'attachment; filename=some_file.pdf'
return response
when i do
return HttpResponse(os.path.join(settings.MEDIA_ROOT, 'QuestionPapers',File_Name))
it returns the absolute path to the file i like to out put but i when use the path in my code as shown above i get the is Errors:
[Errno 2] No such file or directory: 'C:\\Users\\majmaj\\projects\\qbank\\Scripts\\QuestionBank\\media\\QuestionPapers\\filename.pdf'
the directory and file does exist because when i do http://127.0.0.1:8000/media/QuestionPapers/myfile.pdf, the file is opened but i don't know why the open() is not working even when the path to the file is correctly
so one should please look into this and tell me what I'm doing wrong, I'm stuck with this issue and its driving me nuts. this is the fifth day and still no solution
Related
This is a school program to learn how to use file and directory in Python. So to do my best I create a function to open, set it as a variable and close properly my file.
But I got the error of the title:
FileNotFoundError: [Errno 2] No such file or directory: 'codedata.pkl'
def load_db():
""" load data base properly
And get ready for later use
Return:
-------
cd : (list) list of tuples
"""
file = open('codedata.pkl', 'rb')
codedata = pickle.loads(file)
file.close()
return codedata
From the interpreter, this is the line
file = open('codedata.pkl', 'rb')
Which is the problem, but I don't see where is the source of the problem.
Can anyone help me?
Can you check what is the location of the file?
If your file is located at /Users/abc/Desktop/, then the code to open the file on python would be as shown below
file = open('/Users/abc/Desktop/codedata.pkl', 'rb')
codedata = pickle.load(file)
file.close()
You can also check if the file exists at the desired path by doing something like this
import os
filepath = '/Users/abc/Desktop/codedata.pkl'
if os.path.exists(filepath):
file = open('/Users/abc/Desktop/codedata.pkl', 'rb')
codedata = pickle.load(file)
file.close()
else:
print("File not present at desired location")
it happens when you run the script without determining it's the current working directory (example in vs code if you go to Explorer Tape )
You do not work from the same directory that your data.pkl in that's why No file exists
You can know the current directory from getcwd() usually it will be the C/User/.
print(os.getcwd())
filepath=""
if os.path.exists(r"D:\research\StleGAN\karras2019stylegan-ffhq-1024x1024.pkl"):
print("yes")
else:
print("no")
The solution is to open a directory that contains the script or to add the full path.
Issue: Unable to save file in directory (/root/Notion/Image) when using Cron schedule
This is what my code is trying to do:
Check email
Download image attachment
Store in a directory - root/Notion/Image
Retrieve file path
The script is working when I run it manually in Google Cloud terminal. The problem is when I try to schedule it on Cron, it's unable to access the folder to save the file locally.
This is the error when the script failed and require permission:
Traceback (most recent call last):
File "Notion/test.py", line 121, in <module>
path = get_attachments(email.message_from_bytes(msg[0][1]))
File "Notion/test.py", line 47, in get_attachments
with open(filePath, 'wb') as f:
PermissionError: [Errno 13] Permission denied: '/root/Notion/Image/3.jpeg'
This is the code to retrieve attachment from email
def get_attachments(msg):
for part in msg.walk():
if part.get_content_maintype()=='multipart':
continue
if part.get('Content-Disposition') is None:
continue
fileName = part.get_filename()
if bool(fileName):
filePath = os.path.join(attachment_dir, fileName)
with open(filePath, 'wb') as f:
f.write(part.get_payload(decode=True))
return str(filePath)
Resolved:
The problem is that I shouldn't use root directory since it requires permission. I've changed it to home directory instead.
attachment_dir = '/home/dev_thomas_yang/folder_name/folder_name'
For people who needs to check their home direction, simply run this script.
from pathlib import Path
home= str(Path.home())
print(home)
Thanks Triplee for the patience to breakdown my issue despite my sloppy ways of presenting it!
The easiest fix hands down is to change the code so it doesn't try to write to /root. Have it write to the invoking user's home directory instead.
Your question doesn't show the relevant parts of the code, but just change attachment_dir so it's not an absolute path. Maybe separately take care of creating the directory if it doesn't already exist.
import pathlib
# ...
attachment_dir = pathlib.Path("cron/whatever/attachments").mkdir(parents=True, exist_ok=True)
# ...
for loop in circumstances:
get_attachments(something)
A better design altogether would be to have get_attachments accept the directory name as a parameter, so you can make this configurable from the code which calls it. Global variables are a nuisance and cause hard-to-debug problems because they hide information which is important for understanding the code, and tricky to change when you try to debug that code and don't know which parts of the code depend on the old value.
I'm following a simple tutorial on YouTube about how to automatically upload files in S3 using Python, and I'm getting this error shows that:
FileNotFoundError: [WinError 2] The system cannot find the file specified: 'age.csv'
And this does not make sense to me, because files are there. For example, this my code looks like:
client = boto3.client('s3',
aws_access_key_id=access_key,
aws_secret_access_key=secret_access_key)
path = 'C:/Users/User/Desktop/python/projects/AWS-Data-Processing/example_data'
for file in os.listdir(path):
upload_file_bucket = 'my-uploaded-data'
print(file)
if '.txt' in file:
upload_file_key_txt = 'txt/' + str(file)
client.upload_file(file, upload_file_bucket, upload_file_key_txt)
print("txt")
elif '.csv' in file:
upload_file_key_csv = 'csv/' + str(file)
client.upload_file(file, upload_file_bucket, upload_file_key_csv)
print("csv")
And when I comment out the part where it says:
client.upload_file(file, upload_file_bucket, upload_file_key_txt)
it prints out either "txt" or "cvs", and I comment out to just read files such as:
for file in os.listdir(path):
upload_file_bucket = 'my-uploaded-data'
print(file)
Then it successfully prints out the file names. So I don't understand why I get the error of there is no file existing when there is. It sounds contradicting and I need some help to understand this error.
I read a post where I might need to download AWS CLI, so which I did but it didn't help. I'm guessing the problem lies in the function upload_file but I just don't understand how there is no file?
Any advice will be appreciated!
The upload_file function takes a full file path, and not just a name. It cannot guess what is your directory, so you need to prepend it or use a different way of iterating over the files.
Source: https://boto3.amazonaws.com/v1/documentation/api/latest/guide/s3-uploading-files.html
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 trying to read a .json file from within my Flask application using:
def renderblog():
with open(url_for("static", filename="blogs.json")) as blog_file:
data = json.load(blog_file)
However I get the error:
FileNotFoundError: [Errno 2] No such file or directory: '/static/blogs.json'
Now I know for a fact that the directory exists within my project structure, but I have no idea why I'm getting this error. Any ideas? Is there a specific way to retrieve .json in Flask?
You generated a URL path, not a path to the local static folder. Use the app.static_folder attribute instead:
def renderblog():
filename = os.path.join(app.static_folder, 'blogs.json')
with open(filename) as blog_file:
data = json.load(blog_file)