I am trying to convert json to csv and download a file from my flask application. The function does not work correctly, I always get the same csv, even if I delete the json file. Why?
button:
Download
My method:
#app.route("/download/<file_id>")
def get_csv(file_id):
try:
file_id = f"{file_id}"
filename_jsonl = f"{file_id}.jsonl"
filename_csv = f"{file_id}.csv"
file_id = ''
with open(filename_jsonl, 'r') as f:
for line in f.read():
file_id += line
file_id = [json.loads(item + '\n}') for item in file_id.split('}\n')[0:-1]]
with open(filename_csv, 'a') as f:
writer = csv.DictWriter(f, file_id[0].keys(), delimiter=";")
writer.writeheader()
for profile in file_id:
writer.writerow(profile)
return send_from_directory(directory='', filename=filename_csv, as_attachment=True)
except FileNotFoundError:
abort(404)
The problem you are having is that the first generated file has been cached.
Official documentation says that send_from_directory() send a file from a given directory with send_file(). send_file() sets the cache_timeout option.
You must configure this option to disable caching, like this:
return send_from_directory(directory='', filename=filename_csv, as_attachment=True, cache_timeout=0)
#app.route('/download')
def download():
return send_from_directory('static', 'files/cheat_sheet.pdf')
Note: First parameter give it the directory name like static if your file is inside static (the file only could be in the project directory),
and for the second parameter write the right path of the file. The file will be automatically downloaded, if the route got download.
Related
I'm using this to connect to Azure File Share and upload a file. I would like to chose what extension file will have, but I can't. I got an error shown below. If I remove .txt everything works fine. Is there a way to specify file extension while uploading it?
Error:
Exception: ResourceNotFoundError: The specified parent path does not exist.
Code:
def main(blobin: func.InputStream):
file_client = ShareFileClient.from_connection_string(conn_str="<con_string>",
share_name="data-storage",
file_path="outgoing/file.txt")
f = open('/home/temp.txt', 'w+')
f.write(blobin.read().decode('utf-8'))
f.close()
# Operation on file here
f = open('/home/temp.txt', 'rb')
string_to_upload = f.read()
f.close()
file_client.upload_file(string_to_upload)
I believe the reason you're getting this error is because outgoing folder doesn't exist in your file service share. I took your code and ran it with and without extension and in both situation I got the same error.
Then I created a folder and tried to upload the file and I was able to successfully do so.
Here's the final code I used:
from azure.storage.fileshare import ShareFileClient, ShareDirectoryClient
conn_string = "DefaultEndpointsProtocol=https;AccountName=myaccountname;AccountKey=myaccountkey;EndpointSuffix=core.windows.net"
share_directory_client = ShareDirectoryClient.from_connection_string(conn_str=conn_string,
share_name="data-storage",
directory_path="outgoing")
file_client = ShareFileClient.from_connection_string(conn_str=conn_string,
share_name="data-storage",
file_path="outgoing/file.txt")
# Create folder first.
# This operation will fail if the directory already exists.
print "creating directory..."
share_directory_client.create_directory()
print "directory created successfully..."
# Operation on file here
f = open('D:\\temp\\test.txt', 'rb')
string_to_upload = f.read()
f.close()
#Upload file
print "uploading file..."
file_client.upload_file(string_to_upload)
print "file uploaded successfully..."
I have written some code to read the contents from a specific url:
import requests
import os
def read_doc(doc_ID):
filename = doc_ID + ".txt"
if not os.path.exists(filename):
my_url = encode_url(doc_ID) #this is a call to another function that would encode the url
my_response = requests.get(my_url)
if my_response.status_code == requests.codes.ok:
return my_response.text
return None
This checks if there's a file named doc_ID.txt (where doc_ID could be any name provided). And if there's no such file, it would read the contents from a specific url and would return them. What I would like to do is to store those returned contents in a file called doc_ID.txt. That is, I would like to finish my function by creating a new file in case it didn't exist at the beginning.
How can I do that? I tried this:
my_text = my_response.text
output = os.rename(my_text, filename)
return output
but then, the actual contents of the file would become the name of the file and I would get an error saying the filename is too long.
So the issue I think I'm seeing is that you want to put the contents of your request's response into the file, rather than naming the file with the contents. The code below should create a file with the filename you want, and insert the text from your response!
import requests
import os
def read_doc(doc_ID):
filename = doc_ID + ".txt"
if not os.path.exists(filename):
my_url = encode_url(doc_ID) #this is a call to another function that would encode the url
my_response = requests.get(my_url)
if my_response.status_code == requests.codes.ok:
with open(filename, "w") as file:
file.write(my_response.text)
return file
return None
To write the response text to the file, you can simply use python file object, https://docs.python.org/3/tutorial/inputoutput.html#reading-and-writing-files
with open(filename, "w") as file:
file.write(my_text)
I have a Flask view that generates data and saves it as a CSV file with Pandas, then displays the data. A second view serves the generated file. I want to remove the file after it is downloaded. My current code raises a permission error, maybe because after_request deletes the file before it is served with send_from_directory. How can I delete a file after serving it?
def process_data(data)
tempname = str(uuid4()) + '.csv'
data['text'].to_csv('samo/static/temp/{}'.format(tempname))
return file
#projects.route('/getcsv/<file>')
def getcsv(file):
#after_this_request
def cleanup(response):
os.remove('samo/static/temp/' + file)
return response
return send_from_directory(directory=cwd + '/samo/static/temp/', filename=file, as_attachment=True)
after_request runs after the view returns but before the response is sent. Sending a file may use a streaming response; if you delete it before it's read fully you can run into errors.
This is mostly an issue on Windows, other platforms can mark a file deleted and keep it around until it not being accessed. However, it may still be useful to only delete the file once you're sure it's been sent, regardless of platform.
Read the file into memory and serve it, so that's it's not being read when you delete it later. In case the file is too big to read into memory, use a generator to serve it then delete it.
#app.route('/download_and_remove/<filename>')
def download_and_remove(filename):
path = os.path.join(current_app.instance_path, filename)
def generate():
with open(path) as f:
yield from f
os.remove(path)
r = current_app.response_class(generate(), mimetype='text/csv')
r.headers.set('Content-Disposition', 'attachment', filename='data.csv')
return r
I have a Flask view that generates data and saves it as a CSV file with Pandas, then displays the data. A second view serves the generated file. I want to remove the file after it is downloaded. My current code raises a permission error, maybe because after_request deletes the file before it is served with send_from_directory. How can I delete a file after serving it?
def process_data(data)
tempname = str(uuid4()) + '.csv'
data['text'].to_csv('samo/static/temp/{}'.format(tempname))
return file
#projects.route('/getcsv/<file>')
def getcsv(file):
#after_this_request
def cleanup(response):
os.remove('samo/static/temp/' + file)
return response
return send_from_directory(directory=cwd + '/samo/static/temp/', filename=file, as_attachment=True)
after_request runs after the view returns but before the response is sent. Sending a file may use a streaming response; if you delete it before it's read fully you can run into errors.
This is mostly an issue on Windows, other platforms can mark a file deleted and keep it around until it not being accessed. However, it may still be useful to only delete the file once you're sure it's been sent, regardless of platform.
Read the file into memory and serve it, so that's it's not being read when you delete it later. In case the file is too big to read into memory, use a generator to serve it then delete it.
#app.route('/download_and_remove/<filename>')
def download_and_remove(filename):
path = os.path.join(current_app.instance_path, filename)
def generate():
with open(path) as f:
yield from f
os.remove(path)
r = current_app.response_class(generate(), mimetype='text/csv')
r.headers.set('Content-Disposition', 'attachment', filename='data.csv')
return r
Am using Bottle to create an upload API. The script below is able to upload a file to a directory but got two issues which I need to address. One is how can I avoid loading the whole file to memory the other is how to set a maximum size for upload file?
Is it possible to continuously read the file and dump what has been read to file till the upload is complete? the upload.save(file_path, overwrite=False, chunk_size=1024) function seems to load the whole file into memory. In the tutorial, they have pointed out that using .read() is dangerous.
from bottle import Bottle, request, run, response, route, default_app, static_file
app = Bottle()
#route('/upload', method='POST')
def upload_file():
function_name = sys._getframe().f_code.co_name
try:
upload = request.files.get("upload_file")
if not upload:
return "Nothing to upload"
else:
#Get file_name and the extension
file_name, ext = os.path.splitext(upload.filename)
if ext in ('.exe', '.msi', '.py'):
return "File extension not allowed."
#Determine folder to save the upload
save_folder = "/tmp/{folder}".format(folder='external_files')
if not os.path.exists(save_folder):
os.makedirs(save_folder)
#Determine file_path
file_path = "{path}/{time_now}_{file}".\
format(path=save_folder, file=upload.filename, timestamp=time_now)
#Save the upload to file in chunks
upload.save(file_path, overwrite=False, chunk_size=1024)
return "File successfully saved {0}{1} to '{2}'.".format(file_name, ext, save_folder)
except KeyboardInterrupt:
logger.info('%s: ' %(function_name), "Someone pressed CNRL + C")
except:
logger.error('%s: ' %(function_name), exc_info=True)
print("Exception occurred111. Location: %s" %(function_name))
finally:
pass
if __name__ == '__main__':
run(host="localhost", port=8080, reloader=True, debug=True)
else:
application = default_app()
I also tried doing a file.write but same case. File is getting read to memory and it hangs the machine.
file_to_write = open("%s" %(output_file_path), "wb")
while True:
datachunk = upload.file.read(1024)
if not datachunk:
break
file_to_write.write(datachunk)
Related to this, I've seen the property MEMFILE_MAX where several SO posts claim one could set the maximum file upload size. I've tried setting it but it seems not to have any effect as all files no matter the size are going through.
Note that I want to be able to receive office document which could be plain with their extensions or zipped with a password.
Using Python3.4 and bottle 0.12.7
Basically, you want to call upload.read(1024) in a loop. Something like this (untested):
with open(file_path, 'wb') as dest:
chunk = upload.read(1024)
while chunk:
dest.write(chunk)
chunk = upload.read(1024)
(Do not call open on upload; it's already open for you.)
This SO answer includes more example sof how to read a large file without "slurping" it.