在Django中使用来自RESTFUL API的数据的正确方法 [英] Proper way to consume data from RESTFUL API in django

问题描述

我正在尝试学习django，因此当我有最新的解决方案时，我不确定它是否遵循django中的最佳做法。我想在我的网站上显示来自Web api的信息。假设api网址如下：

I'm trying to learn django so while I have a current solution I'm not sure if it follows best practices in django. I would like to display information from a web api on my website. Let's say the api url is as follows:

http://api.example.com/books?author=edwards&year=2009

Thsis将返回Edwards在2009年写的书的清单。格式：

Thsis would return a list of books by Edwards written in the year 2009. Returned in the following format:

{'results':
             [
                {
                   'title':'Book 1',
                   'Author':'Edwards Man',
                   'Year':2009
                },
                {
                   'title':'Book 2',
                   'Author':'Edwards Man',
                   'Year':2009}
           ]
}

当前，我正在如下视图中使用API​​：

Currently I am consuming the API in my views file as follows:

class BooksPage(generic.TemplateView):
    def get(self,request):
        r = requests.get('http://api.example.com/books?author=edwards&year=2009')
        books = r.json()
        books_list = {'books':books['results']}
        return render(request,'books.html',books_list)

通常，我们从models.py文件中的数据库中获取数据，但是我不确定是否应该在models.py或views.py中获取此API数据。如果应该在models.py中，可以有人提供如何执行此操作的示例吗？我为stackoverflow专门编写了上面的示例，因此任何错误都完全是在此处编写的结果。

Normally, we grab data from the database in the models.py file, but I am unsure if I should be grabbing this API data in models.py or views.py. If it should be in models.py, can someone provide an example of how to do this? I wrote the above example sepecifically for stackoverflow, so any bugs are purely a result of writing it here.

推荐答案

我喜欢这种方法将这种逻辑放在单独的服务层（services.py）中；从Django ORM的角度来看，您呈现的数据绝不是模型，它不仅仅是简单的视图逻辑。干净的封装可确保您执行以下操作：控制与支持服务的接口（即，使其看起来像Python API还是带有参数的URL），添加诸如缓存的增强功能（如@sobolevn所述），单独测试API，等等。

I like the approach of putting that kind of logic in a separate service layer (services.py); the data you are rendering is quite not a "model" in the Django ORM sense, and it's more than simple "view" logic. A clean encapsulation ensures you can do things like control the interface to the backing service (i.e., make it look like a Python API vs. URL with parameters), add enhancements such as caching, as @sobolevn mentioned, test the API in isolation, etc.

所以我建议一个简单的 services.py ，看起来像这样：

So I'd suggest a simple services.py, that looks something like this:

def get_books(year, author):
    url = 'http://api.example.com/books' 
    params = {'year': year, 'author': author}
    r = requests.get(url, params=params)
    books = r.json()
    books_list = {'books':books['results']}
    return books_list

请注意参数通过（使用请求程序包的功能）。

Note how the parameters get passed (using a capability of the requests package).

然后在视图中。 py ：

import services
class BooksPage(generic.TemplateView):
    def get(self,request):
        books_list = services.get_books('2009', 'edwards')
        return render(request,'books.html',books_list)

另请参见：

• Separation of business logic and data access in django

这篇关于在Django中使用来自RESTFUL API的数据的正确方法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！