从tastypie uri获取模型对象？ [英] Get model object from tastypie uri?

问题描述

例如：

如果您在python中将uri作为字符串，您如何获取该字符串的模型对象？

解决方案

您可以使用 get_via_uri ，但是@Zakum提到，这将适用您可能不想要的授权。所以挖掘这个方法的源码，我们看到我们可以像这样解析URI：

  from django.core.urlresolvers import解析，get_script_prefix 
 
 def get_pk_from_uri（uri）：
 prefix = get_script_prefix（）
 chomped_uri = uri 
 
如果前缀和chomped_uri.startswith（前缀） ：
 chomped_uri = chomped_uri [len（prefix）-1：] 
 
 try：
 view，args，kwargs = resolve（chomped_uri）
除了Resolver404：
 raise NotFound（提供的URL'％s'不是有效资源的链接。％uri）
 
 return kwargs ['pk'] 
  

如果您的Django应用程序位于Web服务器的根目录（即 get_script_prefix（）=='/'），那么你可以简化为：

  view，args，kwargs = resolve（uri）
 pk = kwargs ['pk'] 
  

How do you get the model object of a tastypie modelresource from it's uri?

for example:

if you were given the uri as a string in python, how do you get the model object of that string?

解决方案

You can use get_via_uri, but as @Zakum mentions, that will apply authorization, which you probably don't want. So digging into the source for that method we see that we can resolve the URI like this:

from django.core.urlresolvers import resolve, get_script_prefix

def get_pk_from_uri(uri):
    prefix = get_script_prefix()
    chomped_uri = uri

    if prefix and chomped_uri.startswith(prefix):
        chomped_uri = chomped_uri[len(prefix)-1:]

    try:
        view, args, kwargs = resolve(chomped_uri)
    except Resolver404:
        raise NotFound("The URL provided '%s' was not a link to a valid resource." % uri)

    return kwargs['pk']

If your Django application is located at the root of the webserver (i.e. get_script_prefix() == '/') then you can simplify this down to:

view, args, kwargs = resolve(uri)
pk = kwargs['pk']

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