Question

我正在使用django 1.5.5和python 2.6

我有这个模型

class Site(models.Model):
    url = models.CharField(max_length = 512)

我想自定义模型，以便具有'www'前缀的网址和不会返回相同对象的网站的网站。
因此，如果我有一个url ='http://foo.com'的网站，则以下所有内容都将返回相同的对象

mysite = Site.objects.get(url__iexact='http://foo.com')
mysite = Site.objects.get(url__iexact='http://www.foo.com')
mysite = Site.objects.filter(url__iexact='http://foo.com')
mysite = Site.objects.filter(url__iexact='http://www.foo.com')

我正在考虑制作类似的方法。

@classmethod
def get_site(cls,url):
    # search for site with url = url
    if url.startswith('http://www'):
         # search without www
    else:
         # search with www
    return site

但我确信有更好的方法可以让我继续使用objects.get和objects.filter

UPDATE：

根据Gonzalo Delgado的建议，我做了一个自定义模型经理

这是我的代码

def url_variants(url):
prefixes = ['http://www.','https://www.','http://','https://',]  # order must be from longest to shortest
for prefix in prefixes:
    if url.startswith(prefix):
        url = url[len(prefix):]
        break
return [ prefix+url for prefix in prefixes]

class SiteManager(models.Manager):
    def filter(self, *args, **kwargs):            
        if 'url' in kwargs:
            variants = url_variants(url)
            # in order to chain '__in' and '__iexact' Q is needed
            q_list = [Q(url__iexact=n) for n in variants]
            q_list = reduce(lambda a, b: a | b, q_list)
            args =  (q_list,) + args            
            kwargs.pop("url", None) # remove original Field lookups
        return super(SiteManager, self).filter(*args, **kwargs)

这很好，现在唯一的问题是，如果我使用任何类型的Field查找，那么它不会使用新的逻辑。
所以任何类型的url__in，url__contains等都不起作用我确信有一种比实现django中可用的每个字段查找更好的方法。

Answer 1

您需要创建custom manager并扩展get并过滤methods：

class SiteManager(models.Manager):
    def get(self, *args, **kwargs):
        if 'url' in kwargs:
            # handle 'www' prefix here and update kwargs['url'] accordingly
        return super(SiteManager, self).get(*args, **kwargs)


    def filter(self, *args, **kwargs):
        if 'url' in kwargs:
            # handle 'www' prefix here and update kwargs['url'] accordingly
        return super(SiteManager, self).filter(*args, **kwargs)

class Site(models.Model):
    url = models.CharField(max_length = 512)

    objects = SiteManager()

Answer 2

我想我找到了解决方案。它并不完美，但我认为这对我来说已经足够了它基于Gonzalo Delgado的答案。

这是我的代码：

from django.db import models
from django.db.models import Q

def url_variants(urls):
    if isinstance(urls, basestring): # if it is not a list then make a list
        urls = [urls]
    variants = []
    for url in urls:
        prefixes = ['http://www.','https://www.','http://','https://',]  # order must be from longest to shortest
        for prefix in prefixes:
            if url.startswith(prefix):
                url = url[len(prefix):]
                break
        variants += [ prefix+url for prefix in prefixes]
    return variants

def variants_filter(variants,lookup):        
    q_list = [Q(**{ lookup: n}) for n in variants]
    q_list = reduce(lambda a, b: a | b, q_list)
    return (q_list,)

class SiteManager(models.Manager):
    def filter(self, *args, **kwargs):
    # find keys that contain 'url'        
        for key in kwargs:
            if key.startswith('url'):                
                variants = url_variants(kwargs[key])                                            
                args = variants_filter(variants, key.replace('__in','')) + args # if there is an '__in' then remove it we already have list support            
                kwargs.pop(key, None)                
                break            
        return super(SiteManager, self).filter(*args, **kwargs)

Django如何按字段覆盖模型匹配

UPDATE：

2 个答案: