Having following code for POST request with file attached:
file = open(file_name, 'rb')
files = {'file': file}
import requests
res = requests.request(method='POST', url=port_url, files=files, data=params)
file here of type <class '_io.BufferedReader'>.
Now I am trying to impliment similar with file body received by me from incoming email attachment as bytes array (want to avoid storing file to local disk):
file = file_body
files = {'file': file}
import requests
res = requests.request(method='POST', url=port_url, files=files, data=params)
and it doesn't work for some reason. Not sure what is wrong actually but external service (CloudConvert) I am trying to send unable to recognize file correctly.
Here file of type <class 'bytes'>.
Tried to convert it to BytesIo or BufferedReader as suggested here without luck.
Could someone help me to post file taken from bytes array with requests.request?
I've been trying to get JSON response out of the following code, but it's throwing an error
from requests import get
import json
url = "https://api.wheretheiss.at/v1/satellites/25544"
response = get(url)
for item in response:
print(item, response[item])
I wanna print the JSON in the following format:
https://i.stack.imgur.com/ZWMbr.png
your code works well to get the correct respone from the url. The problem is that you should parse the raw respone as the json format. Please refer to the following code:
from requests import get
import json
url = "https://api.wheretheiss.at/v1/satellites/25544"
response = get(url)
# dump response into json
result = response.content.decode('utf-8')
result = json.loads(result)
print(result) # print raw json
# print as key-value pairs
for item in result:
print("{0} : {1}".format(item, result[item]))
The output is like following:
Example output
I think you're misunderstanding what the response contains.
Your initial code is good.
from requests import get
import json
url = "https://api.wheretheiss.at/v1/satellites/25544"
response = get(url)
You now have a response. see here for more info.
In this response you have status_code, content, encoding, and the response content as text. it also contains other information about the request. eg we don't want to continue if the request failed
You can simply parse the json by calling json.loads on the decoded response via response.text
parsed_json = json.loads(response.text)
or
parsed_json = response.json()
for item in json_:
print(item, json_[item])
I hope it helps. it is also duplicate to HTTP requests and JSON parsing in Python
I have two servers:
First server send some files to second server and then get the zip file. I need get this zip file from response.
My first server:
files = {'file': ...}
ceb_response = requests.post(ceb, files=files)
My second server respond:
return HttpResponse(open('db.zip', 'rb'))
I need to save db.zip file in first server, how can I get the file from <Response [200]> and save file locally on first server?
The response from your second server carry your zip file by byte[].
You can save the file in the 1st server like:
fileName = '/xxx/db.zip'
with open(fileName, 'wb') as f:
f.write(ceb_response.content)
I found this answer:
import io, requests, pyzipper
r = requests.post(ceb + 'sync', files=files)
with pyzipper.AESZipFile(io.BytesIO(r.content)) as zf:
zf.setpassword(b'1111')
print(zf)
I'm performing a simple task of uploading a file using Python requests library. I searched Stack Overflow and no one seemed to have the same problem, namely, that the file is not received by the server:
import requests
url='http://nesssi.cacr.caltech.edu/cgi-bin/getmulticonedb_release2.cgi/post'
files={'files': open('file.txt','rb')}
values={'upload_file' : 'file.txt' , 'DB':'photcat' , 'OUT':'csv' , 'SHORT':'short'}
r=requests.post(url,files=files,data=values)
I'm filling the value of 'upload_file' keyword with my filename, because if I leave it blank, it says
Error - You must select a file to upload!
And now I get
File file.txt of size bytes is uploaded successfully!
Query service results: There were 0 lines.
Which comes up only if the file is empty. So I'm stuck as to how to send my file successfully. I know that the file works because if I go to this website and manually fill in the form it returns a nice list of matched objects, which is what I'm after. I'd really appreciate all hints.
Some other threads related (but not answering my problem):
Send file using POST from a Python script
http://docs.python-requests.org/en/latest/user/quickstart/#response-content
Uploading files using requests and send extra data
http://docs.python-requests.org/en/latest/user/advanced/#body-content-workflow
If upload_file is meant to be the file, use:
files = {'upload_file': open('file.txt','rb')}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
and requests will send a multi-part form POST body with the upload_file field set to the contents of the file.txt file.
The filename will be included in the mime header for the specific field:
>>> import requests
>>> open('file.txt', 'wb') # create an empty demo file
<_io.BufferedWriter name='file.txt'>
>>> files = {'upload_file': open('file.txt', 'rb')}
>>> print(requests.Request('POST', 'http://example.com', files=files).prepare().body.decode('ascii'))
--c226ce13d09842658ffbd31e0563c6bd
Content-Disposition: form-data; name="upload_file"; filename="file.txt"
--c226ce13d09842658ffbd31e0563c6bd--
Note the filename="file.txt" parameter.
You can use a tuple for the files mapping value, with between 2 and 4 elements, if you need more control. The first element is the filename, followed by the contents, and an optional content-type header value and an optional mapping of additional headers:
files = {'upload_file': ('foobar.txt', open('file.txt','rb'), 'text/x-spam')}
This sets an alternative filename and content type, leaving out the optional headers.
If you are meaning the whole POST body to be taken from a file (with no other fields specified), then don't use the files parameter, just post the file directly as data. You then may want to set a Content-Type header too, as none will be set otherwise. See Python requests - POST data from a file.
(2018) the new python requests library has simplified this process, we can use the 'files' variable to signal that we want to upload a multipart-encoded file
url = 'http://httpbin.org/post'
files = {'file': open('report.xls', 'rb')}
r = requests.post(url, files=files)
r.text
Client Upload
If you want to upload a single file with Python requests library, then requests lib supports streaming uploads, which allow you to send large files or streams without reading into memory.
with open('massive-body', 'rb') as f:
requests.post('http://some.url/streamed', data=f)
Server Side
Then store the file on the server.py side such that save the stream into file without loading into the memory. Following is an example with using Flask file uploads.
#app.route("/upload", methods=['POST'])
def upload_file():
from werkzeug.datastructures import FileStorage
FileStorage(request.stream).save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
return 'OK', 200
Or use werkzeug Form Data Parsing as mentioned in a fix for the issue of "large file uploads eating up memory" in order to avoid using memory inefficiently on large files upload (s.t. 22 GiB file in ~60 seconds. Memory usage is constant at about 13 MiB.).
#app.route("/upload", methods=['POST'])
def upload_file():
def custom_stream_factory(total_content_length, filename, content_type, content_length=None):
import tempfile
tmpfile = tempfile.NamedTemporaryFile('wb+', prefix='flaskapp', suffix='.nc')
app.logger.info("start receiving file ... filename => " + str(tmpfile.name))
return tmpfile
import werkzeug, flask
stream, form, files = werkzeug.formparser.parse_form_data(flask.request.environ, stream_factory=custom_stream_factory)
for fil in files.values():
app.logger.info(" ".join(["saved form name", fil.name, "submitted as", fil.filename, "to temporary file", fil.stream.name]))
# Do whatever with stored file at `fil.stream.name`
return 'OK', 200
You can send any file via post api while calling the API just need to mention files={'any_key': fobj}
import requests
import json
url = "https://request-url.com"
headers = {"Content-Type": "application/json; charset=utf-8"}
with open(filepath, 'rb') as fobj:
response = requests.post(url, headers=headers, files={'file': fobj})
print("Status Code", response.status_code)
print("JSON Response ", response.json())
#martijn-pieters answer is correct, however I wanted to add a bit of context to data= and also to the other side, in the Flask server, in the case where you are trying to upload files and a JSON.
From the request side, this works as Martijn describes:
files = {'upload_file': open('file.txt','rb')}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
However, on the Flask side (the receiving webserver on the other side of this POST), I had to use form
#app.route("/sftp-upload", methods=["POST"])
def upload_file():
if request.method == "POST":
# the mimetype here isnt application/json
# see here: https://stackoverflow.com/questions/20001229/how-to-get-posted-json-in-flask
body = request.form
print(body) # <- immutable dict
body = request.get_json() will return nothing. body = request.get_data() will return a blob containing lots of things like the filename etc.
Here's the bad part: on the client side, changing data={} to json={} results in this server not being able to read the KV pairs! As in, this will result in a {} body above:
r = requests.post(url, files=files, json=values). # No!
This is bad because the server does not have control over how the user formats the request; and json= is going to be the habbit of requests users.
Upload:
with open('file.txt', 'rb') as f:
files = {'upload_file': f.read()}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
Download (Django):
with open('file.txt', 'wb') as f:
f.write(request.FILES['upload_file'].file.read())
Regarding the answers given so far, there was always something missing that prevented it to work on my side. So let me show you what worked for me:
import json
import os
import requests
API_ENDPOINT = "http://localhost:80"
access_token = "sdfJHKsdfjJKHKJsdfJKHJKysdfJKHsdfJKHs" # TODO: get fresh Token here
def upload_engagement_file(filepath):
url = API_ENDPOINT + "/api/files" # add any URL parameters if needed
hdr = {"Authorization": "Bearer %s" % access_token}
with open(filepath, "rb") as fobj:
file_obj = fobj.read()
file_basename = os.path.basename(filepath)
file_to_upload = {"file": (str(file_basename), file_obj)}
finfo = {"fullPath": filepath}
upload_response = requests.post(url, headers=hdr, files=file_to_upload, data=finfo)
fobj.close()
# print("Status Code ", upload_response.status_code)
# print("JSON Response ", upload_response.json())
return upload_response
Note that requests.post(...) needs
a url parameter, containing the full URL of the API endpoint you're calling, using the API_ENDPOINT, assuming we have an http://localhost:8000/api/files endpoint to POST a file
a headers parameter, containing at least the authorization (bearer token)
a files parameter taking the name of the file plus the entire file content
a data parameter taking just the path and file name
Installation required (console):
pip install requests
What you get back from the function call is a response object containing a status code and also the full error message in JSON format. The commented print statements at the end of upload_engagement_file are showing you how you can access them.
Note: Some useful additional information about the requests library can be found here
Some may need to upload via a put request and this is slightly different that posting data. It is important to understand how the server expects the data in order to form a valid request. A frequent source of confusion is sending multipart-form data when it isn't accepted. This example uses basic auth and updates an image via a put request.
url = 'foobar.com/api/image-1'
basic = requests.auth.HTTPBasicAuth('someuser', 'password123')
# Setting the appropriate header is important and will vary based
# on what you upload
headers = {'Content-Type': 'image/png'}
with open('image-1.png', 'rb') as img_1:
r = requests.put(url, auth=basic, data=img_1, headers=headers)
While the requests library makes working with http requests a lot easier, some of its magic and convenience obscures just how to craft more nuanced requests.
In Ubuntu you can apply this way,
to save file at some location (temporary) and then open and send it to API
path = default_storage.save('static/tmp/' + f1.name, ContentFile(f1.read()))
path12 = os.path.join(os.getcwd(), "static/tmp/" + f1.name)
data={} #can be anything u want to pass along with File
file1 = open(path12, 'rb')
header = {"Content-Disposition": "attachment; filename=" + f1.name, "Authorization": "JWT " + token}
res= requests.post(url,data,header)
i have very large POSt data (over 100 MB) with one cookie, now i want to send it to a server through Python, the POSt request is in a file like this:
a=true&b=random&c=2222&d=pppp
This is my following code which only sends Cookies but not the POST content.
import requests
import sys
count = len(sys.argv)
if count < 3:
print 'usage a.py FILE URL LOGFILE'
else:
url = sys.argv[2]
data= {'file': open(sys.argv[1], 'rb')}
cookies = {'session':'testsession'}
r = requests.get(url, data=data, cookies=cookies)
f = open(sys.argv[3], 'w')
f.write(r.text)
f.close()
The code takes File which has POSt data, then the URL to send it , then the OUTPUT file where the response is to be stored.
Note: I am not trying to upload a file but to send the post content which is inside a file.
Firstly you should be using requests.post. Secondly if you want to post just the data inside the file you need to read the contents of the file and parse it to a dict since this is the format that requests.post expects data to come in.
Example: (Note: Just showing the relevant parts)
import urlparse
import requests
import sys
with open(sys.argv[1], 'rb') as f:
data = urlparse.parse_qs('&'.join(f.read().splitlines()))
r = requests.post(url, data=data, cookies=cookies)