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Flask: send_from_directory

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.

Getting error "Process cannot access file" while deleting file in Python

I am working on a project where I have many json files stored in a directory. I need to read all the files and check there data. If 'sessionisfalse` in any of the files, I need to delete the file. Below is the code:
files = os.listdir(config_files_path)
for file in files:
file_path = config_files_path + '//' + file
f = open(file_path)
data = json.load(f)
if not data['session']:
# delete this file
os.remove(file_path)
In the above code, I am getting the list of all the files. Then iterating over each file and reading its content in data. if not data['session'], I need to delete that file. But doing so I am getting process cannot access the file as its being used by another process. Is there any way we can delete the file. Please help. Thanks

You need to close the file before deleting it.
Use f.close() before os.remove(file_path) statement

The error is raised because the file is still open. You must close it before trying to delete it. So, after you have loaded the data, do f.close(). Like this:
files = os.listdir(config_files_path)
for file in files:
file_path = config_files_path + '//' + file
f = open(file_path)
data = json.load(f)
f.close() # <-- close the file
if not data['session']:
# delete this file
os.remove(file_path)

Close already open csv in Python

Is there a way for Python to close that the file is already open file.
Or at the very least display a popup that file is open or a custom written error message popup for permission error.
As to avoid:
PermissionError: [Errno 13] Permission denied: 'C:\\zf.csv'
I've seen a lot of solutions that open a file then close it through python. But in my case. Lets say I left my csv open and then tried to run the job.
How can I make it so it closes the currently opened csv?
I've tried the below variations but none seem to work as they expect that I have already opened the csv at an earlier point through python. I suspect I'm over complicating this.
f = 'C:\\zf.csv'
file.close()
AttributeError: 'str' object has no attribute 'close'
This gives an error as there is no reference to opening of file but simply strings.
Or even..
theFile = open(f)
file_content = theFile.read()
# do whatever you need to do
theFile.close()
As well as:
fileobj=open('C:\\zf.csv',"wb+")
if not fileobj.closed:
print("file is already opened")
How do I close an already open csv?
The only workaround I can think of would be to add a messagebox, though I can't seem to get it to detect the file.
filename = "C:\\zf.csv"
if not os.access(filename, os.W_OK):
print("Write access not permitted on %s" % filename)
messagebox.showinfo("Title", "Close your CSV")

Try using a with context, which will manage the close (__exit__) operation smoothly at the end of the context:
with open(...) as theFile:
file_content = theFile.read()

You can also try to copy the file to a temporary file, and open/close/remove it at will. It requires that you have read access to the original, though.
In this example I have a file "test.txt" that is write-only (chmod 444) and it throws a "Permission denied" error if I try writing to it directly. I copy it to a temporary file that has "777" rights so that I can do what I want with it:
import tempfile, shutil, os
def create_temporary_copy(path):
temp_dir = tempfile.gettempdir()
temp_path = os.path.join(temp_dir, 'temp_file_name')
os.chmod(temp_path, 0o777); # give full access to the tempfile so we can copy
shutil.copy2(path, temp_path) # copy the original into the temp one
os.chmod(temp_path, 0o777); # replace permissions from the original file
return temp_path
path = "./test.txt" # original file
copy_path = create_temporary_copy(path) # temp copy
with open(copy_path, "w") as g: # can do what I want with it
g.write("TEST\n")

f = open("C:/Users/amol/Downloads/result.csv", "r")
print(f.readlines()) #just to check file is open
f.close()
# here you can add above print statement to check if file is closed or not. I am using python 3.5

Creating an on-the-fly zip file from string content for AWS Lambda in Python

I have a Python script that creates a Lambda script in AWS along with all the policies and triggers. I use python boto3 library for that. I create the zip file for the lambda as on-the-fly rather than uploading a static zip file from the hard drive. I use this simple code from below to create my zip file. It creates the zip file without any problems and my python code uploads this zip file as a lambda script and I can view my lambda script in the AWS without any problems. But when I run my lambda script it gives me the module not found error even though I can clearly see that both the module name and the file name does exist and is view-able.
Unable to import module 'xxxx': No module named xxxx
In the file system I double click that zip file that was created by this code and see that the content is created and everything looks normal.
If I bypass zipping on the fly and create the zip statically using WinZip and let the rest of the Python & boto3 script upload this file then it works just fine.
def CreateLambdaZip(self, fileName, fileContent):
with zipfile.ZipFile('Lambda/' + fileName + '.zip', 'w') as myzipc:
myzipc.writestr( fileName + '.py', fileContent)
myzipc.close()
It kinda looks like for the zip file I'm skipping some special headers that is needed by Aws Lambda. Is there such thing? Because in the file system the zip file that is created by Python code and the other one that is created by WinZip are exactly the same. So I know there's nothing wrong with the lambda script.
Update: I'm uploading the zip file using the below code that reads the zip file which was created using the above snippet.
with open('Lambda/'+ fileName +'.zip', 'rb') as zipFile:
func = boto3.client("Lambda").create_function(
FunctionName=lambdaFunction,
Runtime='python2.7',
Role=role['Role']['Arn'],
Handler= fileName + "." + functionName,
Description=description,
Timeout=10,
MemorySize=256,
Publish=True,
Code={'ZipFile': zipFile.read()},
)
When I use zipFile.read() I get 2 different headers for the same content when I zip it using WinZip and when I zip it using Python's module.
Zip file that's created programmatically using Python
b'PK\x03\x04\x14\x00\x00\x00\x00\x00\xe4~\x01IO\x96J=Z\x07\x00\x00Z\x07\x00\x00\x19\x00\x00\x00schedule-ec2-snapshots.pyimport json\nimport boto3\nimport time\nfrom datetime import date, timedelta\n\nprint(\'Loading scheduled EC2 backup actions\')\n\ndef create_snapshots(event, context):\n """\n Lambda function that executes daily snapshots for the instances that
and zipfile created by WinZip
b'PK\x03\x04\x14\x00\x02\x00\x08\x004X\xfcH\x88\x1f\xce\xb5&\x03\x00\x00b\x07\x00\x00\x19\x00\x00\x00schedule-ec2-snapshots.py\x8dU]k\xdb#\x10|7\xf4?,\nA\x12qL\xda\x06B\r~I\x93Bh\x9b\x87&\xf4E\x15\xe1\xac[\xdb\xd7HwBw2\t\xc1\xff\xbd{+\xeb\xcb.\xb4\n\xc4\xba\xdb\xd1\xec\xce\xdc\xae\xa4\x8a\xd2T\x0e~[\xa3\'\xaa\xb9_\x1ag>\xb6\x0b\xa7\n\x9c\xac*S\x80\x14\x0e\xfd\n\xf6\x11\xbf\x9er\\b\xee\xc4dRVJ\xbb(\xfcf\x84Tz\r6\xdb\xa0\xacs\x94p\xfb\xf9\x03,E\xf6\\\x97

With the info above I was able to start the in-memory solution. The deployment of that zip file worked but I could not use the resulting function. Got error:
Unable to import module '<function-name>': No module named <function-name>
I got it to work by specifying the file permissions.
I then use the in-mem-zip to create an AWS lambda function.
Setup:
file_map is a dictionary of full_path->file_bytes.
files is a list of full_paths
def create_lambda_function(function_name, desc, role, handler, file_map, files)
zip_contents = create_in_mem_zip_archive(file_map, files)
result = lambda_code.create_function(
FunctionName=function_name,
Runtime="python2.7",
Description=desc,
Role=role,
Handler=handler,
Code={'ZipFile': zip_contents},
)
return result
def create_in_mem_zip_archive(file_map, files):
buf = io.BytesIO()
logger.info("Building zip file: " + str(files))
with zipfile.ZipFile(buf, 'w', zipfile.ZIP_DEFLATED) as zfh:
for file_name in files:
file_blob = file_map.get(file_name)
if file_blob is None:
logger.error("Missing file {} from files".format(file_name))
continue
try:
info = zipfile.ZipInfo(file_name)
info.date_time = time.localtime()
info.compress_type = zipfile.ZIP_DEFLATED
info.external_attr = 0777 << 16L # give full access
# info.external_attr = 0644 << 16L # -r-wr--r--
# info.external_attr = 0755 << 16L # -rwxr-xr-x
zfh.writestr(info, file_blob)
except Exception as ex:
logger.info("Error reading file: " + file_name + ", error: " + ex.message)
buf.seek(0)
return buf.read()

I have experienced the exactly same problem you have. My solution is do NOT use on the fly zip file. Create a real zip file and add real file into it, and it just works. You can do that even in the lambda environment, by create filepath like "/tmp/yourfile.txt" you can create temp real file when lambda execute.

How to NOT load whole file into memory during upload

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.