Question

我正在尝试使用与Django在管理页面中使用的相同搜索引擎来在我的自定义视图中显示结果。

我发现this website建议使用以下代码：

from django.db.models import Q
from django.db.models.query import QuerySet
import operator

def django_admin_keyword_search(model, keywords, base_qs=None):
    """Search according to fields defined in Admin's search_fields"""
    if not keywords: 
        return []
    fields = model._meta.admin.search_fields

    qs = QuerySet(model)

    for keyword in keywords:
        or_queries = [Q(**{'%s__icontains' % field: keyword}) for field in fields]
        if base_qs is None:
            other_qs = QuerySet(model)
        else:
            other_qs = base_qs
        if qs._select_related:
            other_qs = other_qs.select_related()
        other_qs = other_qs.filter(reduce(operator.or_, or_queries))
        qs = qs & other_qs

    return qs

def do_keyword_search(model, query, base_qs=None):
    return django_admin_keyword_search(model, query.split(' '),
                                       base_qs=base_qs)

但是如果没有在这段代码中获得属性错误（'选项'对象没有属性'admin'），我就无法运行：fields = model._meta.admin.search_fields

修改使用管理默认搜索引擎我做错了什么？

这是使用默认管理搜索引擎的其他任何方式吗？

Answer 1

我设法构建了自己的通用搜索，就像管理搜索一样。课程如下：

from django.db.models import Q
from django.db.models.query import QuerySet
import operator

def django_admin_keyword_search(model, keywords, search_fields):
    """Search according to fields defined in Admin's search_fields"""
    all_queries = None

    for keyword in keywords.split(' '):  #breaks query_string into 'Foo' and 'Bar'
        keyword_query = None

        for field in search_fields:
            each_query = Q(**{field+'__icontains':keyword})

            if not keyword_query:
                keyword_query = each_query
            else:
                keyword_query = keyword_query | each_query

        if not all_queries:
            all_queries = keyword_query
        else:
            all_queries = all_queries & keyword_query

    result_set = model.objects.filter(all_queries).distinct()

    return result_set

享受!!

在我自己的视图中使用Django Admin搜索引擎

1 个答案: