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How to save image which sent via flask send_file

I have this code for server
#app.route('/get', methods=['GET'])
def get():
return send_file("token.jpg", attachment_filename=("token.jpg"), mimetype='image/jpg')
and this code for getting response
r = requests.get(url + '/get')
And i need to save file from response to hard drive. But i cant use r.files. What i need to do in these situation?

Assuming the get request is valid. You can use use Python's built in function open, to open a file in binary mode and write the returned content to disk. Example below.
file_content = requests.get('http://yoururl/get')
save_file = open("sample_image.png", "wb")
save_file.write(file_content.content)
save_file.close()
As you can see, to write the image to disk, we use open, and write the returned content to 'sample_image.png'. Since your server-side code seems to be returning only one file, the example above should work for you.

You can set the stream parameter and extract the filename from the HTTP headers. Then the raw data from the undecoded body can be read and saved chunk by chunk.
import os
import re
import requests
resp = requests.get('http://127.0.0.1:5000/get', stream=True)
name = re.findall('filename=(.+)', resp.headers['Content-Disposition'])[0]
dest = os.path.join(os.path.expanduser('~'), name)
with open(dest, 'wb') as fp:
while True:
chunk = resp.raw.read(1024)
if not chunk: break
fp.write(chunk)

Change twitter banner from url

How would I go by changing the twitter banner using an image from url using tweepy library: https://github.com/tweepy/tweepy/blob/v2.3.0/tweepy/api.py#L392
So far I got this and it returns:
def banner(self):
url = 'https://blog.snappa.com/wp-content/uploads/2019/01/Twitter-Header-Size.png'
file = requests.get(url)
self.api.update_profile_banner(filename=file.content)
ValueError: stat: embedded null character in path
It seems like filename requires an image to be downloaded. Anyway to process this without downloading the image and then removing it?

Looking at library's code you can do what you want.
def update_profile_banner(self, filename, *args, **kargs):
f = kargs.pop('file', None)
So what you need to do is supply the filename and the file kwarg:
filename = url.split('/')[-1]
self.api.update_profile_banner(filename, file=file.content)

import tempfile
def banner():
url = 'file_url'
file = requests.get(url)
temp = tempfile.NamedTemporaryFile(suffix=".png")
try:
temp.write(file.content)
self.api.update_profile_banner(filename=temp.name)
finally:
temp.close()

Importing and naming a remote file

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)

How to download file from ftp?

I'm scripting an install script in python.
How do I download file from ftp in python?
Operating system -- Windows XP - if that makes a difference.

from urllib2 import urlopen
req = urlopen('ftp://ftp.gnu.org/README')
Then you can use req.read() to load the file content into a variable or do anything else with it, or shutil.copyfileobj to save the content to a disk without loading it to memory.

Here's a code snippet I'm currently using.
import mimetypes
import os
import urllib2
import urlparse
def filename_from_url(url):
return os.path.basename(urlparse.urlsplit(url)[2])
def download_file(url):
"""Create an urllib2 request and return the request plus some useful info"""
name = filename_from_url(url)
r = urllib2.urlopen(urllib2.Request(url))
info = r.info()
if 'Content-Disposition' in info:
# If the response has Content-Disposition, we take filename from it
name = info['Content-Disposition'].split('filename=')[1]
if name[0] == '"' or name[0] == "'":
name = name[1:-1]
elif r.geturl() != url:
# if we were redirected, take the filename from the final url
name = filename_from_url(r.geturl())
content_type = None
if 'Content-Type' in info:
content_type = info['Content-Type'].split(';')[0]
# Try to guess missing info
if not name and not content_type:
name = 'unknown'
elif not name:
name = 'unknown' + mimetypes.guess_extension(content_type) or ''
elif not content_type:
content_type = mimetypes.guess_type(name)[0]
return r, name, content_type
Usage:
req, filename, content_type = download_file('http://some.url')
Then you can use req as a file-like object and e.g. use shutil.copyfileobj() to copy the file contents into a local file. If the MIME type doesn't matter simply remove that part of the code.
Since you seem to be lazy, here's code downloading the file directly to a local file:
import shutil
def download_file_locally(url, dest):
req, filename, content_type = download_file(url)
if dest.endswith('/'):
dest = os.path.join(dest, filename)
with open(dest, 'wb') as f:
shutil.copyfileobj(req, f)
req.close()
This method is smart enough to use the filename sent by the server if you specify a path ending with a slash, otherwise it uses the destination you specified.

Use ftplib
Code Sample from the documentation:
>>> from ftplib import FTP
>>> ftp = FTP('ftp.cwi.nl') # connect to host, default port
>>> ftp.login() # user anonymous, passwd anonymous#
>>> ftp.retrlines('LIST') # list directory contents
total 24418
drwxrwsr-x 5 ftp-usr pdmaint 1536 Mar 20 09:48 .
dr-xr-srwt 105 ftp-usr pdmaint 1536 Mar 21 14:32 ..
-rw-r--r-- 1 ftp-usr pdmaint 5305 Mar 20 09:48 INDEX
.
.
.
>>> ftp.retrbinary('RETR README', open('README', 'wb').write)
'226 Transfer complete.'
>>> ftp.quit()

from urllib.request import urlopen
try:
req = urlopen('ftp://ftp.expasy.org/databases/enzyme/enzclass.txt')
except:
print ("Error")

How to download a file using python in a 'smarter' way?

I need to download several files via http in Python.
The most obvious way to do it is just using urllib2:
import urllib2
u = urllib2.urlopen('http://server.com/file.html')
localFile = open('file.html', 'w')
localFile.write(u.read())
localFile.close()
But I'll have to deal with the URLs that are nasty in some way, say like this: http://server.com/!Run.aspx/someoddtext/somemore?id=121&m=pdf. When downloaded via the browser, the file has a human-readable name, ie. accounts.pdf.
Is there any way to handle that in python, so I don't need to know the file names and hardcode them into my script?

Download scripts like that tend to push a header telling the user-agent what to name the file:
Content-Disposition: attachment; filename="the filename.ext"
If you can grab that header, you can get the proper filename.
There's another thread that has a little bit of code to offer up for Content-Disposition-grabbing.
remotefile = urllib2.urlopen('http://example.com/somefile.zip')
remotefile.info()['Content-Disposition']

Based on comments and #Oli's anwser, I made a solution like this:
from os.path import basename
from urlparse import urlsplit
def url2name(url):
return basename(urlsplit(url)[2])
def download(url, localFileName = None):
localName = url2name(url)
req = urllib2.Request(url)
r = urllib2.urlopen(req)
if r.info().has_key('Content-Disposition'):
# If the response has Content-Disposition, we take file name from it
localName = r.info()['Content-Disposition'].split('filename=')[1]
if localName[0] == '"' or localName[0] == "'":
localName = localName[1:-1]
elif r.url != url:
# if we were redirected, the real file name we take from the final URL
localName = url2name(r.url)
if localFileName:
# we can force to save the file as specified name
localName = localFileName
f = open(localName, 'wb')
f.write(r.read())
f.close()
It takes file name from Content-Disposition; if it's not present, uses filename from the URL (if redirection happened, the final URL is taken into account).

Combining much of the above, here is a more pythonic solution:
import urllib2
import shutil
import urlparse
import os
def download(url, fileName=None):
def getFileName(url,openUrl):
if 'Content-Disposition' in openUrl.info():
# If the response has Content-Disposition, try to get filename from it
cd = dict(map(
lambda x: x.strip().split('=') if '=' in x else (x.strip(),''),
openUrl.info()['Content-Disposition'].split(';')))
if 'filename' in cd:
filename = cd['filename'].strip("\"'")
if filename: return filename
# if no filename was found above, parse it out of the final URL.
return os.path.basename(urlparse.urlsplit(openUrl.url)[2])
r = urllib2.urlopen(urllib2.Request(url))
try:
fileName = fileName or getFileName(url,r)
with open(fileName, 'wb') as f:
shutil.copyfileobj(r,f)
finally:
r.close()

2 Kender:
if localName[0] == '"' or localName[0] == "'":
localName = localName[1:-1]
it is not safe -- web server can pass wrong formatted name as ["file.ext] or [file.ext'] or even be empty and localName[0] will raise exception.
Correct code can looks like this:
localName = localName.replace('"', '').replace("'", "")
if localName == '':
localName = SOME_DEFAULT_FILE_NAME

Using wget:
custom_file_name = "/custom/path/custom_name.ext"
wget.download(url, custom_file_name)
Using urlretrieve:
urllib.urlretrieve(url, custom_file_name)
urlretrieve also creates the directory structure if not exists.

You need to look into 'Content-Disposition' header, see the solution by kender.
How to download a file using python in a 'smarter' way?
Posting his solution modified with a capability to specify an output folder:
from os.path import basename
import os
from urllib.parse import urlsplit
import urllib.request
def url2name(url):
return basename(urlsplit(url)[2])
def download(url, out_path):
localName = url2name(url)
req = urllib.request.Request(url)
r = urllib.request.urlopen(req)
if r.info().has_key('Content-Disposition'):
# If the response has Content-Disposition, we take file name from it
localName = r.info()['Content-Disposition'].split('filename=')[1]
if localName[0] == '"' or localName[0] == "'":
localName = localName[1:-1]
elif r.url != url:
# if we were redirected, the real file name we take from the final URL
localName = url2name(r.url)
localName = os.path.join(out_path, localName)
f = open(localName, 'wb')
f.write(r.read())
f.close()
download("https://example.com/demofile", '/home/username/tmp')
I have just updated the answer of kender for python3