Tweak docs

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Tom Christie 2013-07-03 13:08:43 +01:00
parent 5427d90fa4
commit 7398464b39

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@ -16,10 +16,15 @@ Extends [Django's existing `RequestFactory` class][requestfactory].
The `APIRequestFactory` class supports an almost identical API to Django's standard `RequestFactory` class. This means the that standard `.get()`, `.post()`, `.put()`, `.patch()`, `.delete()`, `.head()` and `.options()` methods are all available.
#### Using the format arguments
# Using the standard RequestFactory API to create a form POST request
factory = APIRequestFactory()
request = factory.post('/notes/', {'title': 'new idea'})
#### Using the `format` argument
Methods which create a request body, such as `post`, `put` and `patch`, include a `format` argument, which make it easy to generate requests using a content type other than multipart form data. For example:
# Create a JSON POST request
factory = APIRequestFactory()
request = factory.post('/notes/', {'title': 'new idea'}, format='json')
@ -27,7 +32,9 @@ By default the available formats are `'multipart'` and `'json'`. For compatibil
To support a wider set of request formats, or change the default format, [see the configuration section][configuration].
If you need to explictly encode the request body, you can do so by explicitly setting the `content_type` flag. For example:
#### Explicitly encoding the request body
If you need to explictly encode the request body, you can do so by setting the `content_type` flag. For example:
request = factory.post('/notes/', json.dumps({'title': 'new idea'}), content_type='application/json')
@ -40,7 +47,7 @@ For example, using `APIRequestFactory`, you can make a form PUT request like so:
factory = APIRequestFactory()
request = factory.put('/notes/547/', {'title': 'remember to email dave'})
Using Django's `Factory`, you'd need to explicitly encode the data yourself:
Using Django's `RequestFactory`, you'd need to explicitly encode the data yourself:
factory = RequestFactory()
data = {'title': 'remember to email dave'}