我有一个基本的Django模型,例如:
class Business(models.Model): name = models.CharField(max_length=200, unique=True) email = models.EmailField() phone = models.CharField(max_length=40, blank=True, null=True) description = models.TextField(max_length=500)
我需要对上述模型执行复杂的查询,例如:
qset = ( Q(name__icontains=query) | Q(description__icontains=query) | Q(email__icontains=query) ) results = Business.objects.filter(qset).distinct()
我已经尝试过以下使用法式饼的运气:
def build_filters(self, filters=None): if filters is None: filters = {} orm_filters = super(BusinessResource, self).build_filters(filters) if('query' in filters): query = filters['query'] print query qset = ( Q(name__icontains=query) | Q(description__icontains=query) | Q(email__icontains=query) ) results = Business.objects.filter(qset).distinct() orm_filters = {'query__icontains': results} return orm_filters
在梅花类的Meta类中,我将过滤设置为:
filtering = { 'name: ALL, 'description': ALL, 'email': ALL, 'query': ['icontains',], }
关于如何解决这个问题有什么想法吗?
你走在正确的轨道上。但是,build_filters应该将资源查找转换为ORM查找。
build_filters
默认实现将查询关键字基于__key_bits,值对进行拆分,然后尝试查找所查找资源与其等效ORM之间的映射。
__key_bits
你的代码不应仅在此处构建过滤器。这是一个改进的固定版本:
def build_filters(self, filters=None): if filters is None: filters = {} orm_filters = super(BusinessResource, self).build_filters(filters) if('query' in filters): query = filters['query'] qset = ( Q(name__icontains=query) | Q(description__icontains=query) | Q(email__icontains=query) ) orm_filters.update({'custom': qset}) return orm_filters def apply_filters(self, request, applicable_filters): if 'custom' in applicable_filters: custom = applicable_filters.pop('custom') else: custom = None semi_filtered = super(BusinessResource, self).apply_filters(request, applicable_filters) return semi_filtered.filter(custom) if custom else semi_filtered
因为你使用的是Q对象,所以标准apply_filters方法不够聪明,无法应用你的自定义过滤器键(因为没有键),但是你可以快速覆盖它并添加一个名为“ custom”的特殊过滤器。这样,你build_filters可以找到一个合适的过滤器,构造它的含义,并将其作为自定义传递给apply_filters,后者将直接应用它,而不是尝试从字典中将其值作为一个项目拆包。
apply_filters