django haystack autocomplete

Question

我正在尝试使用django-haystack(现在大约2天),我已经有了基本的入门示例工作,它看起来非常令人印象深刻.我现在想在haystack上尝试自动完成功能.

http://readthedocs.org/docs/django-haystack/en/v1.2.4/autocomplete.html

第一部分似乎很好:"设置数据"似乎很简单.但是,我不确定需要在哪里写"执行查询":即我应该在哪个视图中包含:

from haystack.query import SearchQuerySet
sqs = SearchQuerySet().filter(content_auto=request.GET.get('q', ''))

我目前的urls.py很简单,设置如下:

urlpatterns = patterns('',
    # Examples:
    ('^hello/$', hello),
    (r'^$', hello), 
    (r'^search/', include('haystack.urls')),
    # url(r'^$', 'mysite.views.home', name='home'),
    # url(r'^mysite/', include('mysite.foo.urls')),

    # Uncomment the admin/doc line below to enable admin documentation:
    #url(r'^admin/doc/', include('django.contrib.admindocs.urls')),

    # Uncomment the next line to enable the admin:
    url(r'^admin/', include(admin.site.urls)),
)

看下面的博客:

http://tech.agilitynerd.com/haystack-search-result-ordering-and-pre-rende

我希望有类似的东西:

url(r'^search/', SearchView(load_all=False,searchqueryset=sqs),name='haystack_search'), 

但是,应该在哪里指定sqs？在urls.py或views.py？我试过两个,但他们给我一个名称错误"请求"在sqs语句中找不到.

Answer 1

通常这会在你自己的 haystack url 中，haystack.urls在你当前的 urls.py 中指向默认 url。创建一个文件 haystack_urls.py 并在 urls.py 中添加它，例如

url(r'^search/', include('yourproject.haystack_urls')),

在该文件中，您可以添加自定义代码，例如

from haystack.views import SearchView
from haystack.query import SearchQuerySet

sqs = SearchQuerySet() # edit remove line that was incorret

urlpatterns = patterns('haystack.views',
    url(r'^&', SearchView(load_all=False,searchqueryset=sqs),name='haystack_search'), 
)

要包装请求的视图，请尝试类似的操作

class SearchWithRequest(SearchView):

    __name__ = 'SearchWithRequest'

    def build_form(self, form_kwargs=None):
        if form_kwargs is None:
            form_kwargs = {}

        if self.searchqueryset is None:
            sqs = SearchQuerySet().filter(content_auto=self.request.GET.get('q', ''))
            form_kwargs['searchqueryset'] = sqs

        return super(SearchWithRequest, self).build_form(form_kwargs)