Related
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 am having the problem that the requests library for some reason is making the payload bigger and causing issues. I enabled http logging, and in in the output I can see the content length being 50569, not 50349, as the actual file size should indicate.
send: b'POST /api/1/images HTTP/1.1\r\nHost: localhost:8000\r\nUser-Agent: python-requests/2.21.0\r\nAccept-Encoding: gzip, deflate\r\nAc
cept: application/json\r\nConnection: keep-alive\r\nAuthorization: Bearer 28956340ba9c7e25b49085b4d273522b\r\ncontent-type: image/png\r\n
Content-Length: 50569\r\n\r\n'
send: b'--ac9e15d6d3aa3a77506c2daccca2ee47\r\nContent-Disposition: form-data; name="0007-Alternate-Lateral-Pulldown_back-STEP1"; filename
="0007-Alternate-Lateral-Pulldown_back-STEP1.png"\r\n\r\n\x89PNG\r\n\x1a\n\x00\x00\x00\rIHDR\x00\x00\x00\xf0\x00\x00\x00\xf0\x08\x06\x00\
x00\x00>U\xe9\x92\x00\x00\x00\tpHYs\x00\x00\x0b\x13\x00\x00\x0b\x13\x01\x00\x9a\x9c\x18\x00\x00FMiTXtXML:com.adobe.xmp\x00\x00\x00\x00\x0
0<?xpacket begin="\xef\xbb\xbf" id="W5M0MpCehiHzreSzNTczkc9d"?>\n<x:xmpmeta xmlns:x="adobe:ns:meta/" x:xmptk="Adobe XMP Core 5.5-c021 79.
155772, 2014/01/13-19:44:00 ">\n <rdf:RDF xmlns:rdf="http:
Chrome has exactly the same headers when sending this, but the correct content length, so I am assuming this is why my server complains of a invalid image being sent.
This is my code
url = self.server + "/api/1/images";
headers = self.default_headers()
headers['content-type'] = 'image/png'
# neither of these are actually used for anything
filename = os.path.basename(image)
field_name = os.path.splitext(filename)[0]
files = {field_name: (filename, open(image, 'rb'), '')}
# Post image
r = requests.post(url, headers=headers, files=files, timeout = 5.0)
As can be seen, I am using the b flag when opening the file to preserve the binary content, so it should not change.
File size is 50349
$ ls -l 0007-Alternate-Lateral-Pulldown_back-STEP1.png
-rw-rw-r--# 1 carlerik staff 50349 Nov 26 2019 0007-Alternate-Lateral-Pulldown_back-STEP1.png
I used Charles proxy to dig into this and I now have gotten to the bottom of this. There are basically two things to note:
The difference in content length between the request sent by Chrome and Requests is exactly the length of the boundary fields (a multipart form concept) before and after the file + the six CRLF (\r\n) sequences + the Content-Disposition header.
echo '50569-178-36-6' | bc
50349
The boundary field looks like this: --ac9e15d6d3aa3a77506c2daccca2ee47\r\n
You can also see from the HTTP header and body dump that the header is actually in the body and after the boundary field, not as part of the normal headers. This was important and led me on the right path.
The second part of the answer is that the guys that wrote the server API I am interfacing with did not understand/read the HTTP spec for the exact bits they ask for: the Content-Disposition header.
The API docs for .../images state that this header must be present always, as they use (well, used in the past) its content to extract filenames and such. The problem is that the way they use it is not in accordance with how it is intended to be used: in a multipart HTTP request it is part of the body of the HTTP request, describing the part (form field) of the request it precedes.
This is, of course, also how Requests uses it, but I did not have this information before venturing into this abyss, so I was misinformed by the code in the controller that states this in its doc. So I assumed that Requests would put the header in the header section, which it did not, and not the body, which it did. After all, I saw that Chrome "did the right thing", but it turned out that was only because these requests were handcrafted in javascript:
apiService.js
/**
* Upload image
* #param file
* #returns {*}
* #private
*/
_api.postImage = function (file) {
if (typeof file != 'object') {
throw Error('ApiService: Object expected');
}
file.fileName = (/(.*)\./).exec(file.name)[1];
var ContentDisposition = 'form-data; name="' + encodeURI(file.fileName) + '"; filename="' + encodeURI(file.name) + '"';
return Upload.http({
url: Routing.generate('api_post_image'),
headers: {
'Content-Disposition': ContentDisposition,
'Content-Type': file.type
},
data: file
});
};
So the Content-Disposition header here is basically a proprietary header to convey information about the filename, that shares its appearance with the general in-body header from the spec. That means all it takes to fix this is to create a request with a custom body read from file and set this header.
To round this off, this was how it was all simplified down to:
headers = dict()
headers['authorization'] = "<something>"
headers['content-type'] = 'image/png'
with open(image, 'rb') as imagefile:
# POST image
r = requests.post(url, headers=headers, data=imagefile)
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 am using a Flask route as a proxy to download a file, like this:
#esa_handler.route("/data/<int:series>/<int:file_num>", methods=["GET"])
def DownloadRemote(series, file_num):
"""
Downloads the remote files from the ESA.
:param series: 0-20.
:param file_num: File within the series, 0-255
:return: Compressed CSV file.
"""
# if the file is bad.
if series >= 20 and file_num > 110:
return jsonify({"error": "file does not exist."})
url = "http://cdn.gea.esac.esa.int/Gaia/gaia_source/csv/GaiaSource_000-{:03d}-{:03d}.csv.gz".format(series,
file_num)
req = requests.get(url, stream=True)
return Response(stream_with_context(req.iter_content(chunk_size=2048)), content_type=req.headers["content-type"])
It works fine, however, the filename that is presented to the client is whatever the file number that is passed to the endpoint. For example, if I put http://127.0.0.1:5000/esa/data/0/0 to download the very first file, it downloads, but Chrome/Firefox/IE/Edge are offering to save the file with a filename as "0". While there is nothing wrong with that, I would like a better user experience.
How can I intercept the response to proffer a filename based off the URL requested?
This can be done with the Content-Disposition HTTP header. Here you can specify a filename for the newly downloaded file.
This can be added to a Flask Response as follows:
url = "http://cdn.gea.esac.esa.int/Gaia/gaia_source/csv/GaiaSource_000-{:03d}-{:03d}.csv.gz".format(series,
req = requests.get(url, stream=True)
headers = Headers()
headers .add('Content-Type', req.headers["content-type"])
headers .add('Content-Disposition', 'attachment; filename="filename.txt"')
return Response(stream_with_context(req.iter_content(chunk_size=2048)), headers=headers)
Note: The Content-Type was moved into headers for simplicity
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)