Question

我想对我所有的API端点（api/v1/）使用api版本索引。目前，我是通过像这样构造urlpatterns来实现的：

urlpatterns = [
    path('api/v1/units/', include('units.api.urls')),
    path('api/v1/accounts/', include('accounts.api.urls')),
]

是否可以更优雅地组织此活动？理想情况下，我希望它看起来像这样：

apipatterns = [
    'units/', include('units.api.urls'),
    'accounts/', include('accounts.api.urls')
]

urlpatterns = [
    path('api/v1/', include(apipatterns)),
]

Answer 1

Including other URLconfs可以使用from http.server import BaseHTTPRequestHandler, HTTPServer from os import path hostName = "localhost" hostPort = 8080 class RequestHandler(BaseHTTPRequestHandler): dir = path.abspath(path.dirname(__file__)) content_type = 'text/html' def _set_headers(self): self.send_response(200) self.send_header('Content-Type', self.content_type) self.send_header('Content-Length', path.getsize(self.getPath())) self.end_headers() def do_GET(self): self._set_headers() self.wfile.write(self.getContent(self.getPath())) def getPath(self): if self.path == '/': content_path = path.join(dir, 'index.html') else: content_path = path.join(dir, str(self.path)) return content_path def getContent(self, content_path): with open(content_path, mode='r', encoding='utf-8') as f: content = f.read() return bytes(content, 'utf-8') myServer = HTTPServer((hostName, hostPort), RequestHandler) myServer.serve_forever()和path。

在这种情况下，您可以尝试：

include

因此，路由apipatterns = [ path('units/', include('units.api.urls')), path('accounts/', include('accounts.api.urls')) ] urlpatterns = [ path('api/v1/', include(apipatterns)), ]将由api/v1/units/处理，而'units.api.urls'将由api/v1/accounts/处理

我希望这会有所帮助。

Answer 2

您的“理想”网址格式几乎是正确的。这是已更正的变体：

'accounts.api.urls'

这里是documentation中的类似示例。

为django urlpatterns优化API

2 个答案: