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How to convert suds object to xml string

This is a duplicate to this question:
How to convert suds object to xml
But the question has not been answered: "totxt" is not an attribute on the Client class.
Unfortunately I lack of reputation to add comments. So I ask again:
Is there a way to convert a suds object to its xml?
I ask this because I already have a system that consumes wsdl files and sends data to a webservice. But now the customers want to alternatively store the XML as files (to import them later manually). So all I need are 2 methods for writing data: One writes to a webservice (implemented and tested), the other (not implemented yet) writes to files.
If only I could make something like this:
xml_as_string = My_suds_object.to_xml()
The following code is just an example and does not run. And it's not elegant. Doesn't matter. I hope you get the idea what I want to achieve:
I have the function "write_customer_obj_webservice" that works. Now I want to write the function "write_customer_obj_xml_file".
import suds
def get_customer_obj():
wsdl_url = r'file:C:/somepathhere/Customer.wsdl'
service_url = r'http://someiphere/Customer'
c = suds.client.Client(wsdl_url, location=service_url)
customer = c.factory.create("ns0:CustomerType")
return customer
def write_customer_obj_webservice(customer):
wsdl_url = r'file:C:/somepathhere/Customer.wsdl'
service_url = r'http://someiphere/Customer'
c = suds.client.Client(wsdl_url, location=service_url)
response = c.service.save(someparameters, None, None, customer)
return response
def write_customer_obj_xml_file(customer):
output_filename = r'C\temp\testxml'
# The following line is the problem. "to_xml" does not exist and I can't find a way to do it.
xml = customer.to_xml()
fo = open(output_filename, 'a')
try:
fo.write(xml)
except:
raise
else:
response = 'All ok'
finally:
fo.close()
return response
# Get the customer object always from the wsdl.
customer = get_customer_obj()
# Since customer is an object, setting it's attributes is very easy. There are very complex objects in this system.
customer.name = "Doe J."
customer.age = 42
# Write the new customer to a webservice or store it in a file for later proccessing
if later_processing:
response = write_customer_obj_xml_file(customer)
else:
response = write_customer_obj_webservice(customer)

I found a way that works for me. The trick is to create the Client with the option "nosend=True".
In the documentation it says:
nosend - Create the soap envelope but don't send. When specified, method invocation returns a RequestContext instead of sending it.
The RequestContext object has the attribute envelope. This is the XML as string.
Some pseudo code to illustrate:
c = suds.client.Client(url, nosend=True)
customer = c.factory.create("ns0:CustomerType")
customer.name = "Doe J."
customer.age = 42
response = c.service.save(someparameters, None, None, customer)
print response.envelope # This prints the XML string that would have been sent.

You have some issues in write_customer_obj_xml_file function:
Fix bad path:
output_filename = r'C:\temp\test.xml'
The following line is the problem. "to_xml" does not exist and I can't find a way to do it.
What's the type of customer? type(customer)?
xml = customer.to_xml() # to be continued...
Why mode='a'? ('a' => append, 'w' => create + write)
Use a with statement (file context manager).
with open(output_filename, 'w') as fo:
fo.write(xml)
Don't need to return a response string: use an exception manager. The exception to catch can be EnvironmentError.
Analyse
The following call:
customer = c.factory.create("ns0:CustomerType")
Construct a CustomerType on the fly, and return a CustomerType instance customer.
I think you can introspect your customer object, try the following:
vars(customer) # display the object attributes
help(customer) # display an extensive help about your instance
Another way is to try the WSDL URLs by hands, and see the XML results.
You may obtain the full description of your CustomerType object.
And then?
Then, with the attributes list, you can create your own XML. Use an XML template and fill it with the object attributes.
You may also found the magic function (to_xml) which do the job for you. But, not sure the XML format matches your need.

client = Client(url)
client.factory.create('somename')
# The last XML request by client
client.last_sent()
# The last XML response from Web Service
client.last_received()

open('file_path').read() - save contents back to file

I need to upload a file to the server using urllib2. Since I cannot use any external libraries (like requests and others) because I am using OpenOffice python, I needed a simple way to post file data.
So I came with:
post_url = "http://localhost:8000/admin/oo_file_uploader?user_id=%s&file_id=%s" % (user_id, file_id)
file_path = doc.Location.replace('file://', '')
data = urllib.urlencode({"file": open(file_path).read()})
urllib2.urlopen(post_url, data)
which posts something to the server.
I wonder if it is possible to save posted contents back to the file using python/django?

This expands somewhat on #zero323's answer. You will want to make sure that you implement some sort of security to prevent random files being uploaded by unauthorized users, which is what the #file_upload_security decorator is implied to handle.
#file_upload_security
def oo_file_uploader(user_id=None, file_id=None):
if request.method == 'POST':
# Exception handling skipped if get() fails.
user = User.objects.get(id=user_id)
save_to_file = MyFiles.objects.get(id=file_id)
# You will probably want to ensure 'file' is in post data.
file_contents = save_to_file.parse_post_to_content(request.POST['file'])
with open(save_to_file.path_to_file, 'w') as fw:
fw.write(file_contents)

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)

Link generator using django or any python module

I want to generate for my users temporary download link.
Is that ok if i use django to generate link using url patterns?
Could it be correct way to do that. Because can happen that I don't understand some processes how it works. And it will overflow my memory or something else. Some kind of example or tools will be appreciated. Some nginx, apache modules probably?
So, what i wanna to achieve is to make url pattern which depend on user and time. Decript it end return in view a file.

A simple scheme might be to use a hash digest of username and timestamp:
from datetime import datetime
from hashlib import sha1
user = 'bob'
time = datetime.now().isoformat()
plain = user + '\0' + time
token = sha1(plain)
print token.hexdigest()
"1e2c5078bd0de12a79d1a49255a9bff9737aa4a4"
Next you store that token in a memcache with an expiration time. This way any of your webservers can reach it and the token will auto-expire. Finally add a Django url handler for '^download/.+' where the controller just looks up that token in the memcache to determine if the token is valid. You can even store the filename to be downloaded as the token's value in memcache.

Yes it would be ok to allow django to generate the urls. This being exclusive from handling the urls, with urls.py. Typically you don't want django to handle the serving of files see the static file docs[1] about this, so get the notion of using url patterns out of your head.
What you might want to do is generate a random key using a hash, like md5/sha1. Store the file and the key, datetime it's added in the database, create the download directory in a root directory that's available from your webserver like apache or nginx... suggest nginx), Since it's temporary, you'll want to add a cron job that checks if the time since the url was generated has expired, cleans up the file and removes the db entry. This should be a django command for manage.py
Please note this is example code written just for this and not tested! It may not work the way you were planning on achieving this goal, but it works. If you want the dl to be pw protected also, then look into httpbasic auth. you can generate and remove entries on the fly in a httpd.auth file using htpasswd and the subprocess module when you create the link or at registration time.
import hashlib, random, datetime, os, shutil
# model to hold link info. has these fields: key (charfield), filepath (filepathfield)
# datetime (datetimefield), url (charfield), orgpath (filepathfield of the orignal path
# or a foreignkey to the files model.
from models import MyDlLink
# settings.py for the app
from myapp import settings as myapp_settings
# full path and name of file to dl.
def genUrl(filepath):
# create a onetime salt for randomness
salt = ''.join(['{0}'.format(random.randrange(10) for i in range(10)])
key = hashlib('{0}{1}'.format(salt, filepath).hexdigest()
newpath = os.path.join(myapp_settings.DL_ROOT, key)
shutil.copy2(fname, newpath)
newlink = MyDlink()
newlink.key = key
newlink.date = datetime.datetime.now()
newlink.orgpath = filepath
newlink.newpath = newpath
newlink.url = "{0}/{1}/{2}".format(myapp_settings.DL_URL, key, os.path.basename(fname))
newlink.save()
return newlink
# in commands
def check_url_expired():
maxage = datetime.timedelta(days=7)
now = datetime.datetime.now()
for link in MyDlink.objects.all():
if(now - link.date) > maxage:
os.path.remove(link.newpath)
link.delete()
[1] http://docs.djangoproject.com/en/1.2/howto/static-files/

It sounds like you are suggesting using some kind of dynamic url conf.
Why not forget your concerns by simplifying and setting up a single url that captures a large encoded string that depends on user/time?
(r'^download/(?P<encrypted_id>(.*)/$', 'download_file'), # use your own regexp
def download_file(request, encrypted_id):
decrypted = decrypt(encrypted_id)
_file = get_file(decrypted)
return _file
A lot of sites just use a get param too.
www.example.com/download_file/?09248903483o8a908423028a0df8032
If you are concerned about performance, look at the answers in this post: Having Django serve downloadable files
Where the use of the apache x-sendfile module is highlighted.
Another alternative is to simply redirect to the static file served by whatever means from django.

passing blob parameter to django

I keep my images in the DB as blobs:
class MyClass(db.Model):
icon=db.BlobProperty()
Now, I want to send the blob to my HTML like this :
lets say I have myClass as an instance of MyClass
result = """<div img_attr=%s> Bla Bla </div>""" % (myClass.icon)
Some how it doesn't work. Do you have any idea?

You cannot just dump raw image data into your html page. You need to do this in two pieces:
In your html, you need to refer to an image file:
result = "<div>"
"<img src='{0}' />"
"</div>"
.format(MYSITE + MYIMAGEDIR + myClass.name)
Your browser reads the html page, finds out you want to include an image, and goes looking for the image file to include - so it asks your site for something like http://www.myexample.com/images/icon01.jpg
Now, separately, you respond to this second request with the image content, as #anand has shown.

Your code suggests that you are working on Google application engine with Django.
You just need to query the image in your view and return it as http response.
image = myclassObject.icon
response = HttpResponse(image)
response['Content-Type'] = 'image/jpg'
return response

The value stored in the the data store, and returned by appengine with a db.BlobProperty is not the actual bytes of the blob, but rather a BlobKey that is used to reference it. There are two ways you can use that key. You can create a BlobReader to load the bytes of the blob from the BlobStore into your app, or you can craft a response with ServeHandler.send_blob to transfer those bytes to the client.
Doing the second one in Django is a bit of a headache, because ServeHandler doesn't really fit in well with the Django request processing stack. Here's a view that will do it for you without too much trouble:
def get_image_data(request, key, filename):
"serve original uploaded image"
# verify the users' ability to get the requested image
key = urllib.unquote(key)
img = _load_metadata(request, key)
blob = img.data;
blobkey = blob.key()
# and tell google to serve it
response = http.HttpResponse(
content='',
content_type=blob.content_type)
response['X-AppEngine-BlobKey'] = str(blobkey)
return response