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How to compress Excel buffer into ZIP buffer?

I'm trying to compress a Excel BytesIO stream into a ZIP BytesIO stream but ValueError: stat: embedded null character in path is called when I use write(). I used pyzipper and pyminizip but neither worked well.
def __compress_excel__(self, excel_buffer):
zip_buffer = BytesIO()
password = b'password'
zip_buffer = pyzipper.AESZipFile(zip_stream, mode='w')
zip_buffer.setpassword(password)
zip_buffer.write(excel_buffer.getvalue())
return zip_buffer.getvalue()
I want to avoid using a temp file to do this. Regards.
UPDATE
Now thanks to martineau comments, Excel could be compressed now, but I have a new problem is that setpassword() from ZipFile doesn't work. A ZIP is created but when in uncompress it, no password is required.
def __compress_excel__(self, excel_buffer):
zip_stream = BytesIO()
password = b'password'
with ZipFile(zip_stream , mode='w') as zipf:
zipf.setpassword(password)
zipf.writestr('excel.xlsx', excel_buffer.getvalue())
return zip_stream.getvalue()
Regards

how to make urllib.request append to an existing file?

I'm trying to download a load of text in Python and want it all to save to a single file.
The code I'm currently using creates a separate file for each url. It loops through an archive of urls, requests the data and then saves it to its own file.
filename = archive[i]
urllib.request.urlretrieve(url, path + filename + ".pgn")
I've tried using the same filename for each url but it just overwrites the file.
Is there a way to loop through the archive and, rather than saving the data in its own separate file, add each block of text to a single file? Or do I need to just loop through all the files afterwards and concatenate them together?

Python's urlretrive docs says that
If you wish to retrieve a resource via URL and store it in a temporary location, you can do so via the urlretrieve() function
so if you wish to append the retrieved data in one file you have use urlopen for that
Like this :
import urllib.request
filename = "MY_FILE_PATH"
#-----------inside your i loop-------------
with urllib.request.urlopen(url) as response:
data = response.read()
# change your file type according e.g. "ab" for binary file
with open(filename + ".pgn", "a+") as fp: fp.write(str(data))

Note that urlretrieve might become deprecated at some point in the future. So use urlopen instead.
import urllib.request
import shutil
...
filename = archive[i]
with urllib.request.urlopen(url) as response, open(filename, 'ab') as out_file:
shutil.copyfileobj(response, out_file)

How to convert base64 String to image and stored in django model?

I'm converting an image to a Base64 string and sending it from android device to the server. Now, I need to change that string back to an image and save it in the database which is Django model
import base64
imgdata = base64.b64decode(imgstring)
filename = 'some_image.jpg' # I assume you have a way of picking unique filenames
with open(filename, 'wb') as f:
f.write(imgdata)

Use:
image = np.asarray(Image.open(io.BytesIO(imgstring)))

Saving a file in a directory but also being able to print it whenever i needto

I am running Python 3.x. So i have been working on some code for fetching data on currencies names around the world from a currency website to get information which the code is as follows
def _fetch_currencies():
import urllib.request
import json
f = urllib.request.urlopen('http://openexchangerates.org/api/currencies.json')
charset = f.info().get_param('charset', 'utf8')
data = f.read()
decoded = json.loads(data.decode(charset))
dumps = json.dumps(decoded, indent=4)
return dumps
I then need to save it as a file locally but having some issue and cant see where.
Here is the code for saving the currencies:
def save_currencies(_fetch_currencies, filename):
sorted_currencies = sorted(decoded.items())
with open(filename, 'w') as my_csv:
csv_writer = csv.writer(my_csv, delimiter=',')
csv_writer.writerows(sorted_currencies)
They just don't seem to work together apart from when i remove the line ' dumps = json.dumps(decoded, indent=4) ' but i need that line to be able to print the file in text, how do i get around deleting this line and still be able to save and print? How do i also pick where it saves?
Any Help will be great, thank you very much anyone and everyone who answers/reads this.

I may be mistaken, but your "decoded" variable should be declared as global in both functions.

I would actually have _fetch_currencies() return a dictionary, and then I would pass that dictionary on to saved_currencies(currencies_decoded, filename). For example:
def _fetch_currencies():
import urllib.request
import json
f = urllib.request.urlopen('http://openexchangerates.org/api/currencies.json')
charset = f.info().get_param('charset', 'utf8')
data = f.read()
decoded = json.loads(data.decode(charset))
return decoded
def save_currencies(currencies_decoded, filename):
sorted_currencies = sorted(currencies_decoded.items())
with open(filename, 'w') as my_csv:
csv_writer = csv.writer(my_csv, delimiter=',')
csv_writer.writerows(sorted_currencies)
my_currencies_decoded = _fetch_currencies()
save_currencies(my_currencies_decoded, "filename.csv")
Furthermore, if you would like to save your csv file to a certain location in your filesystem, you can import os and use the os.path.join() function and provide it the FULL path. For example, to save your .csv file to a location called "/Documents/Location/Here", you can do:
import os
def save_currencies(currencies_decoded, filename):
sorted_currencies = sorted(currencies_decoded.items())
with open(os.path.join("Documents","Location","Here"), 'w') as my_csv:
csv_writer = csv.writer(my_csv, delimiter=',')
csv_writer.writerows(sorted_currencies)
You can also use a relative path, so that if you're already in directory "Documents", and you'd like to save a file in "/Documents/Location/Here", you can instead just say:
with open(os.path.join("Location", "Here"), 'w') as my_csv:

function for loading both strings and files on disk?

I have a design question. I have a function loadImage() for loading an image file. Now it accepts a string which is a file path. But I also want to be able to load files which are not on physical disk, eg. generated procedurally. I could have it accept a string, but then how could it know the string is not a file path but file data? I could add an extra boolean argument to specify that, but that doesn't sound very clean. Any ideas?
It's something like this now:
def loadImage(filepath):
file = open(filepath, 'rb')
data = file.read()
# do stuff with data
The other version would be
def loadImage(data):
# do stuff with data
How to have this function accept both 'filepath' or 'data' and guess what it is?

You can change your loadImage function to expect an opened file-like object, such as:
def load_image(f):
data = file.read()
... and then have that called from two functions, one of which expects a path and the other a string that contains the data:
from StringIO import StringIO
def load_image_from_path(path):
with open(path, 'rb') as f:
load_image(f)
def load_image_from_string(s):
sio = StringIO(s)
try:
load_image(sio)
finally:
sio.close()

How about just creating two functions, loadImageFromString and loadImageFromFile?

This being Python, you can easily distinguish between a filename and a data string. I would do something like this:
import os.path as P
from StringIO import StringIO
def load_image(im):
fin = None
if P.isfile(im):
fin = open(im, 'rb')
else:
fin = StringIO(im)
# Read from fin like you would from any open file object
Other ways to do it would be a try block instead of using os.path, but the essence of the approach remains the same.