[英]Django how do i override model match by field
我正在使用Django 1.5.5和python 2.6
我有這個模特
class Site(models.Model):
url = models.CharField(max_length = 512)
我想自定義模型,以便帶有url帶有“ www”前綴的站點和不具有該站點的站點將返回相同的對象。
因此,如果我有一個網址為url =' http://foo.com '的網站,則以下所有內容將返回同一對象
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')
我當時想做一個類似的類方法。
@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
但是我確信有更好的方法,這樣我就可以繼續使用objects.get
和objects.filter
根據貢薩洛·德爾加多(Gonzalo Delgado)的建議,我做了一個自定義模型經理
這是我的代碼
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)
這很好,現在唯一的問題是,如果我使用任何類型的字段查找,那么它將不使用新邏輯。
因此任何類型的url__in
, url__contains
等均不起作用。
我確信,有比實現django中可用的每個歸檔查詢更好的方法。
您需要創建一個自定義管理器,並擴展get
和filter 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()
我想我找到了解決方案。 它並不完美,但我認為對我來說已經足夠了。
它基於Gonzalo Delgado的答案。
這是我的代碼:
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)
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