We Keep Coding

Storing response as json in Python requests-cache

I'm using requests-cache to cache http responses in human-readable format.
I've patched requests using the filesystem backend, and the the serializer to json, like so:
import requests_cache
requests_cache.install_cache('example_cache', backend='filesystem', serializer='json')
The responses do get cached as json, but the response's body is encoded (I guess using the cattrs library, as described here).
Is there a way to make requests-cache save responses as-is?

What you want to do makes sense, but it's a bit more complicated than it appears. The response files you see are representations of requests.Response objects. Response._content contains the original bytes received from the server. The wrapper methods and properties like Response.json() and Response.text will then attempt to decode that content. For a Response object to work correctly, it needs to have the original binary response body.
When requests-cache serializes that response as JSON, the binary content is encoded in Base85. That's why you're seeing encoded bytes instead of JSON there. To have everything including the response body saved in JSON, there are a couple options:
Option 1
Make a custom serializer. If you wanted to be able to modify response content and have those changes reflected in responses returned by requests-cache, this would probably be the best way to do it.
This may be become a bit convoluted, because you would have to:
Handle response content that isn't valid JSON, and save as encoded bytes instead
During deserialization, if the content was saved as JSON, convert it back into bytes to recreate the original Response object
It's doable, though. I could try to come up with an example later, if needed.
Option 2
Make a custom backend. It could extend FileCache and FileDict, and copy valid JSON content to a separate file. Here is a working example:
import json
from os.path import splitext
from requests import Response
from requests_cache import CachedSession, FileCache, FileDict
class JSONFileCache(FileCache):
"""Filesystem backend that copies JSON-formatted response content into a separate file
alongside the main response file
"""
def __init__(self, cache_name, **kwargs):
super().__init__(cache_name, **kwargs)
self.responses = JSONFileDict(cache_name, **kwargs)
class JSONFileDict(FileDict):
def __setitem__(self, key: str, value: Response):
super().__setitem__(key, value)
response_path = splitext(self._path(key))[0]
json_path = f'{response_path}_content.json'
# Will handle errors and skip writing if content can't be decoded as JSON
with self._try_io(ignore_errors=True):
content = json.dumps(value.json(), indent=2)
with open(json_path, mode='w') as f:
f.write(content)
Usage example:
custom_backend = JSONFileCache('example_cache', serializer='json')
session = CachedSession(backend=custom_backend)
session.get('https://httpbin.org/get')
After making a request, you will see a pair of files like:
example_cache/680f2a52944ee079.json
example_cache/680f2a52944ee079_content.json
That may not be exactly what you want, but it's the easiest option if you only need to read the response content and don't need to modify it.

Python requests_toolbelt MultipartEncoder filename

Using requests_toolbelt to upload large files in a Multipart form, I have constructed a method below which succeeds in uploading the file, however I cannot access the posted filename. How do I access the filename on the server?
# client-side
file = open('/Volumes/Extra/test/my_video.mpg', 'rb')
payload = MultipartEncoder({file.name: file})
r = requests.post(url, data=payload, headers={'Content-Type': 'application/octet-stream'})
# server-side
#view_config(route_name='remote.agent_upload', renderer='json')
def remote_agent_upload(request):
r = request.response
fs = request.body_file
f = open('/Volumes/Extra/tests2/bar.mpg', 'wb') # wish to use filename here
f.write(fs.read())
fs.close()
f.close()
return r

OK, it looks like you are using the name of the file as the field name. Also, the way that you are doing it, seems like the entire post content is being written to file... Is this the desired outcome? Have you tried to actually play your mpg files after you write them on the server side?
I don't have an HTTP server readily available to test at the moment which automagically gives me a request object, but I am assuming that the request object is a webob.Request object (at least it seems like that is the case, please correct me if I'm wrong)
OK, let me show you my test. (This works on python3.4, not sure what version of Python you are using, but I think it should also work on Python 2.7 - not tested though)
The code in this test is a bit long, but it is heavily commented to help you understand what I did every step of the way. Hopefully, it will give you a better understanding of how HTTP requests and responses work in python with the tools you are using
# My Imports
from requests_toolbelt import MultipartEncoder
from webob import Request
import io
# Create a buffer object that can be read by the MultipartEncoder class
# This works just like an open file object
file = io.BytesIO()
# The file content will be simple for my test.
# But you could just as easily have a multi-megabyte mpg file
# Write the contents to the file
file.write(b'test mpg content')
# Then seek to the beginning of the file so that the
# MultipartEncoder can read it from the beginning
file.seek(0)
# Create the payload
payload = MultipartEncoder(
{
# The name of the file upload field... Not the file name
'uploadedFile': (
# This would be the name of the file
'This is my file.mpg',
# The file handle that is ready to be read from
file,
# The content type of the file
'application/octet-stream'
)
}
)
# To send the file, you would use the requests.post method
# But the content type is not application-octet-stream
# The content type is multipart/form-data; with a boundary string
# Without the proper header type, your server would not be able to
# figure out where the file begins and ends and would think the
# entire post content is the file, which it is not. The post content
# might even contain multiple files
# So, to send your file, you would use:
#
# response = requests.post(url, data=payload, headers={'Content-Type': payload.content_type})
# Instead of sending the payload to the server,
# I am just going to grab the output as it would be sent
# This is because I don't have a server, but I can easily
# re-create the object using this output
postData = payload.to_string()
# Create an input buffer object
# This will be read by our server (our webob.Request object)
inputBuffer = io.BytesIO()
# Write the post data to the input buffer so that the webob.Request object can read it
inputBuffer.write(postData)
# And, once again, seek to 0
inputBuffer.seek(0)
# Create an error buffer so that errors can be written to it if there are any
errorBuffer = io.BytesIO()
# Setup our wsgi environment just like the server would give us
environment = {
'HTTP_HOST': 'localhost:80',
'PATH_INFO': '/index.py',
'QUERY_STRING': '',
'REQUEST_METHOD': 'POST',
'SCRIPT_NAME': '',
'SERVER_NAME': 'localhost',
'SERVER_PORT': '80',
'SERVER_PROTOCOL': 'HTTP/1.0',
'CONTENT_TYPE': payload.content_type,
'wsgi.errors': errorBuffer,
'wsgi.input': inputBuffer,
'wsgi.multiprocess': False,
'wsgi.multithread': False,
'wsgi.run_once': False,
'wsgi.url_scheme': 'http',
'wsgi.version': (1, 0)
}
# Create our request object
# This is the same as your request object and should have all our info for reading
# the file content as well as the file name
request = Request(environment)
# At this point, the request object is the same as what you get on your server
# So, from this point on, you can use the following code to get
# your actual file content as well as your file name from the object
# Our uploaded file is in the POST. And the POST field name is 'uploadedFile'
# Grab our file so that it can be read
uploadedFile = request.POST['uploadedFile']
# To read our content, you can use uploadedFile.file.read()
print(uploadedFile.file.read())
# And to get the file name, you can use uploadedFile.filename
print(uploadedFile.filename)
So, I think this modified code will work for you. (Hopefully)
Again, not tested because I don't actually have a server to test with. And also, I don't know what kind of object your "request" object is on the server side.... OK, here goes:
# client-side
import requests
file = open('/Volumes/Extra/test/my_video.mpg', 'rb')
payload = MultipartEncoder({'uploadedFile': (file.name, file, 'application/octet-stream')})
r = requests.post('http://somewhere/somefile.py', data=payload, headers={'Content-Type': payload.content_type})
# server-side
#view_config(route_name='remote.agent_upload', renderer='json')
def remote_agent_upload(request):
# Write your actual file contents, not the post data which contains multi part boundary
uploadedFile = request.POST['uploadedFile']
fs = uploadedFile.file
# The file name is insecure. What if the file name comes through as '../../../etc/passwd'
# If you don't secure this, you've just wiped your /etc/passwd file and your server is toast
# (assuming the web user has write permission to the /etc/passwd file
# which it shouldn't, but just giving you a worst case scenario)
fileName = uploadedFile.filename
# Secure the fileName here...
# Make sure it doesn't have any slashes or double dots, or illegal characters, etc.
# I'll leave that up to you
# Write the file
f = open('/Volumes/Extra/tests2/' + fileName, 'wb')
f.write(fs.read())

Probably too late for the original OP but may help someone else. This is how I upload a file with accompanying json in a multipart/form-data upload using the MultipartEncoder. When I require a file to uploaded as binary with a single json string as part of a multipart request (so there are just two parts, the file and the json). Note that in creating my request header (it's a custom header as designated by the receiving server) I get the content_type from the encoded object (it usually comes through as multipart/form-data). I'm using simplejson.dumps but you can just use json.dumps I believe.
m = MultipartEncoder([
('json', (None, simplejson.dumps(datapayload), 'text/plain')),
('file', (os.path.basename(file_path), open(file_path, 'rb'), 'text/plain'))],
None, encoding='utf-8')
headers = {'Authorization': 'JwToken' + ' ' + jwt_str, 'content-type': m.content_type}
response = requests.post(uri, headers=headers, data=m, timeout=45, verify=True )
In the file part, the field is called "file", but I use os.path.basename(file_path) to get just the filename from a full file path e.g. c:\temp\mytestfile.txt . It's possible I could just as easily call the file something else (that's not the original name) in this field if I wanted to.

Getting an empty file when served by django

I have the following code for managing file download through django.
def serve_file(request, id):
file = models.X.objects.get(id=id).file #FileField
file.open('rb')
wrapper = FileWrapper(file)
mt = mimetypes.guess_type(file.name)[0]
response = HttpResponse(wrapper, content_type=mt)
import unicodedata, os.path
filename = unicodedata.normalize('NFKD', os.path.basename(file.name)).encode("utf8",'ignore')
filename = filename.replace(' ', '-') #Avoid browser to ignore any char after the space
response['Content-Length'] = file.size
response['Content-Disposition'] = 'attachment; filename={0}'.format(filename)
#print response
return response
Unfortunately, my browser get an empty file when downloading.
The printed response seems correct:
Content-Length: 3906
Content-Type: text/plain
Content-Disposition: attachment; filename=toto.txt
blah blah ....
I have similar code running ok. I don't see what can be the problem. Any idea?
PS: I have tested the solution proposed here and get the same behavior
Update:
Replacing wrapper = FileWrapper(file) by wrapper = file.read() seems to fix the problem
Update: If I comment the print response, I get similar issue:. the file is empty. Only difference: FF detects a 20bytes size. (the file is bigger than this)

File object is an interable, and a generator. It can be read only once before being exausted. Then you have to make a new one, of use a method to start at the begining of the object again (e.g: seek()).
read() returns a string, which can be read multiple times without any problem, this is why it solves your issue.
So just make sure that if you use a file like object, you don't read it twice in a row. E.G: don't print it, then returns it.

From django documentation:
FieldFile.open(mode='rb') Behaves like the standard Python open()
method and opens the file associated with this instance in the mode
specified by mode.
If it works like pythons open then it should return a file-object, and should be used like this:
f = file.open('rb')
wrapper = FileWrapper(f)

How do I handle file upload via PUT request in Django?

I'm implementing a REST-style interface and would like to be able to create (via upload) files via a HTTP PUT request. I would like to create either a TemporaryUploadedFile or a InMemoryUploadedFile which I can then pass to my existing FileField and .save() on the object that is part of the model, thereby storing the file.
I'm not quite sure about how to handle the file upload part. Specifically, this being a put request, I do not have access to request.FILES since it does not exist in a PUT request.
So, some questions:
Can I leverage existing functionality in the HttpRequest class, specifically the part that handles file uploads? I know a direct PUT is not a multipart MIME request, so I don't think so, but it is worth asking.
How can I deduce the mime type of what is being sent? If I've got it right, a PUT body is simply the file without prelude. Do I therefore require that the user specify the mime type in their headers?
How do I extend this to large amounts of data? I don't want to read it all into memory since that is highly inefficient. Ideally I'd do what TemporaryUploadFile and related code does - write it part at a time?
I've taken a look at this code sample which tricks Django into handling PUT as a POST request. If I've got it right though, it'll only handle form encoded data. This is REST, so the best solution would be to not assume form encoded data will exist. However, I'm happy to hear appropriate advice on using mime (not multipart) somehow (but the upload should only contain a single file).
Django 1.3 is acceptable. So I can either do something with request.raw_post_data or request.read() (or alternatively some other better method of access). Any ideas?

Django 1.3 is acceptable. So I can
either do something with
request.raw_post_data or
request.read() (or alternatively some
other better method of access). Any
ideas?
You don't want to be touching request.raw_post_data - that implies reading the entire request body into memory, which if you're talking about file uploads might be a very large amount, so request.read() is the way to go. You can do this with Django <= 1.2 as well, but it means digging around in HttpRequest to figure out the the right way to use the private interfaces, and it's a real drag to then ensure your code will also be compatible with Django >= 1.3.
I'd suggest that what you want to do is to replicate the existing file upload behaviour parts of the MultiPartParser class:
Retrieve the upload handers from request.upload_handlers (Which by default will be MemoryFileUploadHandler & TemporaryFileUploadHandler)
Determine the request's content length (Search of Content-Length in HttpRequest or MultiPartParser to see the right way to do this.)
Determine the uploaded file's filename, either by letting the client specify this using the last path part of the url, or by letting the client specify it in the "filename=" part of the Content-Disposition header.
For each handler, call handler.new_file with the relevant args (mocking up a field name)
Read the request body in chunks using request.read() and calling handler.receive_data_chunk() for each chunk.
For each handler call handler.file_complete(), and if it returns a value, that's the uploaded file.
How can I deduce the mime type of what
is being sent? If I've got it right, a
PUT body is simply the file without
prelude. Do I therefore require that
the user specify the mime type in
their headers?
Either let the client specify it in the Content-Type header, or use python's mimetype module to guess the media type.
I'd be interested to find out how you get on with this - it's something I've been meaning to look into myself, be great if you could comment to let me know how it goes!
Edit by Ninefingers as requested, this is what I did and is based entirely on the above and the django source.
upload_handlers = request.upload_handlers
content_type = str(request.META.get('CONTENT_TYPE', ""))
content_length = int(request.META.get('CONTENT_LENGTH', 0))
if content_type == "":
return HttpResponse(status=400)
if content_length == 0:
# both returned 0
return HttpResponse(status=400)
content_type = content_type.split(";")[0].strip()
try:
charset = content_type.split(";")[1].strip()
except IndexError:
charset = ""
# we can get the file name via the path, we don't actually
file_name = path.split("/")[-1:][0]
field_name = file_name
Since I'm defining the API here, cross browser support isn't a concern. As far as my protocol is concerned, not supplying the correct information is a broken request. I'm in two minds as to whether I want say image/jpeg; charset=binary or if I'm going to allow non-existent charsets. In any case, I'm putting setting Content-Type validly as a client-side responsibility.
Similarly, for my protocol, the file name is passed in. I'm not sure what the field_name parameter is for and the source didn't give many clues.
What happens below is actually much simpler than it looks. You ask each handler if it will handle the raw input. As the author of the above states, you've got MemoryFileUploadHandler & TemporaryFileUploadHandler by default. Well, it turns out MemoryFileUploadHandler will when asked to create a new_file decide whether it will or not handle the file (based on various settings). If it decides it's going to, it throws an exception, otherwise it won't create the file and lets another handler take over.
I'm not sure what the purpose of counters was, but I've kept it from the source. The rest should be straightforward.
counters = [0]*len(upload_handlers)
for handler in upload_handlers:
result = handler.handle_raw_input("",request.META,content_length,"","")
for handler in upload_handlers:
try:
handler.new_file(field_name, file_name,
content_type, content_length, charset)
except StopFutureHandlers:
break
for i, handler in enumerate(upload_handlers):
while True:
chunk = request.read(handler.chunk_size)
if chunk:
handler.receive_data_chunk(chunk, counters[i])
counters[i] += len(chunk)
else:
# no chunk
break
for i, handler in enumerate(upload_handlers):
file_obj = handler.file_complete(counters[i])
if not file_obj:
# some indication this didn't work?
return HttpResponse(status=500)
else:
# handle file obj!

Newer Django versions allow for handling this a lot easier thanks to https://gist.github.com/g00fy-/1161423
I modified the given solution like this:
if request.content_type.startswith('multipart'):
put, files = request.parse_file_upload(request.META, request)
request.FILES.update(files)
request.PUT = put.dict()
else:
request.PUT = QueryDict(request.body).dict()
to be able to access files and other data like in POST. You can remove the calls to .dict() if you want your data to be read-only.

I hit this problem while working with Django 2.2, and was looking for something that just worked for uploading a file via PUT request.
from django.http import QueryDict
from django.http.multipartparser import MultiValueDict
from django.core.files.uploadhandler import (
SkipFile,
StopFutureHandlers,
StopUpload,
)
class PutUploadMiddleware(object):
def __init__(self, get_response):
self.get_response = get_response
def __call__(self, request):
method = request.META.get("REQUEST_METHOD", "").upper()
if method == "PUT":
self.handle_PUT(request)
return self.get_response(request)
def handle_PUT(self, request):
content_type = str(request.META.get("CONTENT_TYPE", ""))
content_length = int(request.META.get("CONTENT_LENGTH", 0))
file_name = request.path.split("/")[-1:][0]
field_name = file_name
content_type_extra = None
if content_type == "":
return HttpResponse(status=400)
if content_length == 0:
# both returned 0
return HttpResponse(status=400)
content_type = content_type.split(";")[0].strip()
try:
charset = content_type.split(";")[1].strip()
except IndexError:
charset = ""
upload_handlers = request.upload_handlers
for handler in upload_handlers:
result = handler.handle_raw_input(
request.body,
request.META,
content_length,
boundary=None,
encoding=None,
)
counters = [0] * len(upload_handlers)
for handler in upload_handlers:
try:
handler.new_file(
field_name,
file_name,
content_type,
content_length,
charset,
content_type_extra,
)
except StopFutureHandlers:
break
for chunk in request:
for i, handler in enumerate(upload_handlers):
chunk_length = len(chunk)
chunk = handler.receive_data_chunk(chunk, counters[i])
counters[i] += chunk_length
if chunk is None:
# Don't continue if the chunk received by
# the handler is None.
break
for i, handler in enumerate(upload_handlers):
file_obj = handler.file_complete(counters[i])
if file_obj:
# If it returns a file object, then set the files dict.
request.FILES.appendlist(file_name, file_obj)
break
any(handler.upload_complete() for handler in upload_handlers)

Am I parsing this HTTP POST request properly?

Let me start off by saying, I'm using the twisted.web framework. Twisted.web's file uploading didn't work like I wanted it to (it only included the file data, and not any other information), cgi.parse_multipart doesn't work like I want it to (same thing, twisted.web uses this function), cgi.FieldStorage didn't work ('cause I'm getting the POST data through twisted, not a CGI interface -- so far as I can tell, FieldStorage tries to get the request via stdin), and twisted.web2 didn't work for me because the use of Deferred confused and infuriated me (too complicated for what I want).
That being said, I decided to try and just parse the HTTP request myself.
Using Chrome, the HTTP request is formed like this:
------WebKitFormBoundary7fouZ8mEjlCe92pq
Content-Disposition: form-data; name="upload_file_nonce"
11b03b61-9252-11df-a357-00266c608adb
------WebKitFormBoundary7fouZ8mEjlCe92pq
Content-Disposition: form-data; name="file"; filename="login.html"
Content-Type: text/html
<!DOCTYPE html>
<html>
<head>
...
------WebKitFormBoundary7fouZ8mEjlCe92pq
Content-Disposition: form-data; name="file"; filename=""
------WebKitFormBoundary7fouZ8mEjlCe92pq--
Is this always how it will be formed? I'm parsing it with regular expressions, like so (pardon the wall of code):
(note, I snipped out most of the code to show only what I thought was relevant (the regular expressions (yeah, nested parentheses), this is an __init__ method (the only method so far) in an Uploads class I built. The full code can be seen in the revision history (I hope I didn't mismatch any parentheses)
if line == "--{0}--".format(boundary):
finished = True
if in_header == True and not line:
in_header = False
if 'type' not in current_file:
ignore_current_file = True
if in_header == True:
m = re.match(
"Content-Disposition: form-data; name=\"(.*?)\"; filename=\"(.*?)\"$", line)
if m:
input_name, current_file['filename'] = m.group(1), m.group(2)
m = re.match("Content-Type: (.*)$", line)
if m:
current_file['type'] = m.group(1)
else:
if 'data' not in current_file:
current_file['data'] = line
else:
current_file['data'] += line
you can see that I start a new "file" dict whenever a boundary is reached. I set in_header to True to say that I'm parsing headers. When I reach a blank line, I switch it to False -- but not before checking if a Content-Type was set for that form value -- if not, I set ignore_current_file since I'm only looking for file uploads.
I know I should be using a library, but I'm sick to death of reading documentation, trying to get different solutions to work in my project, and still having the code look reasonable. I just want to get past this part -- and if parsing an HTTP POST with file uploads is this simple, then I shall stick with that.
Note: this code works perfectly for now, I'm just wondering if it will choke on/spit out requests from certain browsers.

My solution to this Problem was parsing the content with cgi.FieldStorage like:
class Root(Resource):
def render_POST(self, request):
self.headers = request.getAllHeaders()
# For the parsing part look at [PyMOTW by Doug Hellmann][1]
img = cgi.FieldStorage(
fp = request.content,
headers = self.headers,
environ = {'REQUEST_METHOD':'POST',
'CONTENT_TYPE': self.headers['content-type'],
}
)
print img["upl_file"].name, img["upl_file"].filename,
print img["upl_file"].type, img["upl_file"].type
out = open(img["upl_file"].filename, 'wb')
out.write(img["upl_file"].value)
out.close()
request.redirect('/tests')
return ''

You're trying to avoid reading documentation, but I think the best advice is to actually read:
rfc 2388 Returning Values from Forms: multipart/form-data
rfc 1867 Form-based File Upload in HTML
to make sure you don't miss any cases. An easier route might be to use the poster library.

The content-disposition header has no defined order for fields, plus it may contain more fields than just the filename. So your match for filename may fail - there may not even be a filename!
See rfc2183 (edit that's for mail, see rfc1806, rfc2616 and maybe more for http)
Also I would suggest in these kind of regexps to replace every space by \s*, and not to rely on character case.