On Python ≥ 3.5, use pathlib.Path.mkdir
:
from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)
For older versions of Python, I see two answers with good qualities, each with a small flaw, so I will give my take on it:
Try os.path.exists
, and consider os.makedirs
for the creation.
import os
if not os.path.exists(directory):
os.makedirs(directory)
As noted in comments and elsewhere, there's a race condition – if the directory is created between the os.path.exists
and the os.makedirs
calls, the os.makedirs
will fail with an OSError
. Unfortunately, blanket-catching OSError
and continuing is not foolproof, as it will ignore a failure to create the directory due to other factors, such as insufficient permissions, full disk, etc.
One option would be to trap the OSError
and examine the embedded error code (see Is there a cross-platform way of getting information from Python’s OSError):
import os, errno
try:
os.makedirs(directory)
except OSError as e:
if e.errno != errno.EEXIST:
raise
Alternatively, there could be a second os.path.exists
, but suppose another created the directory after the first check, then removed it before the second one – we could still be fooled.
Depending on the application, the danger of concurrent operations may be more or less than the danger posed by other factors such as file permissions. The developer would have to know more about the particular application being developed and its expected environment before choosing an implementation.
Modern versions of Python improve this code quite a bit, both by exposing FileExistsError
(in 3.3+)...
try:
os.makedirs("path/to/directory")
except FileExistsError:
# directory already exists
pass
...and by allowing a keyword argument to os.makedirs
called exist_ok
(in 3.2+).
os.makedirs("path/to/directory", exist_ok=True) # succeeds even if directory exists.
Django-rest-framework works well even without tying it to a model. Your approach sounds ok, but I believe you can trim some of the steps to get everything working.
For example, rest framework comes with a few built-in renderers. Out of the box it can return JSON and XML to the API consumer. You can also enable YAML by just installing the required python module. Django-rest-framework will output any basic object like dict, list and tuple without any extra work on your part.
So basically you only have to create the function or class that takes in arguments, does all of the required calculations and returns its results in a tuple to the REST api view. If JSON and/or XML fits your needs, django-rest-framework will take care of the serialization for you.
You can skip steps 2 and 3 in this case, and just use one class for calculations and one for presentation to the API consumer.
Here are a few snippets may help you out:
Please note that I have not tested this. It's only meant as an example, but it should work :)
The CalcClass:
class CalcClass(object):
def __init__(self, *args, **kw):
# Initialize any variables you need from the input you get
pass
def do_work(self):
# Do some calculations here
# returns a tuple ((1,2,3, ), (4,5,6,))
result = ((1,2,3, ), (4,5,6,)) # final result
return result
The REST view:
from rest_framework.views import APIView
from rest_framework.response import Response
from rest_framework import status
from MyProject.MyApp import CalcClass
class MyRESTView(APIView):
def get(self, request, *args, **kw):
# Process any get params that you may need
# If you don't need to process get params,
# you can skip this part
get_arg1 = request.GET.get('arg1', None)
get_arg2 = request.GET.get('arg2', None)
# Any URL parameters get passed in **kw
myClass = CalcClass(get_arg1, get_arg2, *args, **kw)
result = myClass.do_work()
response = Response(result, status=status.HTTP_200_OK)
return response
Your urls.py:
from MyProject.MyApp.views import MyRESTView
from django.conf.urls.defaults import *
urlpatterns = patterns('',
# this URL passes resource_id in **kw to MyRESTView
url(r'^api/v1.0/resource/(?P<resource_id>\d+)[/]?$', login_required(MyRESTView.as_view()), name='my_rest_view'),
url(r'^api/v1.0/resource[/]?$', login_required(MyRESTView.as_view()), name='my_rest_view'),
)
This code should output a list of lists when you access http://example.com/api/v1.0/resource/?format=json. If using a suffix, you can substitute ?format=json
with .json
. You may also specify the encoding you wish to get back by adding "Content-type"
or "Accept"
to the headers.
[
[
1,
2,
3
],
[
4,
5,
6
]
]
Hope this helps you out.
Best Answer
Take a look at the api view from django-rest-framework-jwt. It's an implementation for creating auth tokens rather than cookie sessions, but your implementation will be similar. See views.py and serializers.py. You can probably use the
serializers.py
unchanged, and just adjust your views to return the right parameters and possibly set the session cookie (can't recall if that's already performed in authentication).