python - 覆盖 Haystack 中的用户查询

Question

我希望用户查询在我的 Django/Haystack app 中按字母顺序排序。例如，如果用户输入“sit”，这将在 Haystack 搜索索引中更改为“ist”。

我相信覆盖 default Haystack forms.py将实现这一目标，但无法使其发挥作用。

#forms.py
from haystack.forms import SearchForm

class WordsSearchForm(SearchForm):

    q_default = forms.CharField(required=False, label=_('Search'),
                            widget=forms.TextInput(attrs={'type': 'search'}))

    q = ''.join(sorted(q_default))

关于如何实现这一目标有什么建议吗？

Answer 1

事实证明，自定义 Haystack 有点棘手。要覆盖用户查询，您需要创建一个 forms.py 文件，更新views.py，并更新urls.py。

#forms.py
from haystack.forms import SearchForm

class WordsSearchForm(SearchForm):

def search(self):
    # added the two lines below
    q = self.cleaned_data['q']
    q = ''.join(sorted(q))

    if not self.is_valid():
        return self.no_query_found()

    if not self.cleaned_data.get('q'):
        return self.no_query_found()

    # default is...
    # sqs = self.searchqueryset.auto_query(self.cleaned_data['q'])
    sqs = self.searchqueryset.auto_query(q)

    if self.load_all:
        sqs = sqs.load_all()

    return sqs


# views.py
from haystack.views import SearchView

from .models import [NAME OF YOUR MODEL]
from .forms import WordsSearchForm

class MySearchView(SearchView):

    form = WordsSearchForm


#urls.py
from haystack.views import SearchView, search_view_factory
from scrabble.views import MySearchView
from APP.forms import WordsSearchForm

urlpatterns = patterns('',
    # replace default haystack url
    # url(r'^search/', include('haystack.urls')),

      url(r'^search/$', search_view_factory(
                       form_class=WordsSearchForm
                       ), name='haystack_search'),
)