Question

我有一个基本的Django模型，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)

我需要在上面的模型上执行一个复杂的查询，如：

qset = (
    Q(name__icontains=query) |
    Q(description__icontains=query) |
    Q(email__icontains=query)
    )
results = Business.objects.filter(qset).distinct()

我使用 tastypie 尝试了以下操作但没有运气：

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

并且在Meta for tastypie中，我将过滤设置为：

filtering = {
        'name: ALL,
        'description': ALL,
        'email': ALL,
        'query': ['icontains',],
    }

我有什么想法可以解决这个问题吗？

由于 - 牛顿

Answer 1

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。

默认实现将基于__的查询关键字拆分为key_bits，值对，然后尝试查找查找的资源与其ORM等效项之间的映射。

您的代码不应该应用仅在其中构建的过滤器。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

由于您使用的是Q对象，因此标准apply_filters方法不够智能，无法应用自定义过滤器键（因为没有），但是您可以快速覆盖它并添加一个名为“custom”的特殊过滤器。在这样做的过程中，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，它将直接应用它而不是尝试将其值从字典中解压缩为项目。

Answer 2

我这样解决了这个问题：

Class MyResource(ModelResource):

  def __init__(self, *args, **kwargs):
    super(MyResource, self).__init__(*args, **kwargs)
    self.q_filters = []

  def build_filters(self, filters=None):
    orm_filters = super(MyResource, self).build_filters(filters)

    q_filter_needed_1 = []
    if "what_im_sending_from_client" in filters:
      if filters["what_im_sending_from_client"] == "my-constraint":
        q_filter_needed_1.append("something to filter")

    if q_filter_needed_1:
      a_new_q_object = Q()
      for item in q_filter_needed:
        a_new_q_object = a_new_q_object & Q(filtering_DB_field__icontains=item)
      self.q_filters.append(a_new_q_object)

  def apply_filters(self, request, applicable_filters):
    filtered = super(MyResource, self).apply_filters(request, applicable_filters)

    if self.q_filters:
      for qf in self.q_filters:
        filtered = filtered.filter(qf)
      self.q_filters = []

    return filtered

这种方法比我见过的其他方法感觉更清晰。

Answer 3

根据astevanovic的回答并清理一下这个想法，以下内容应该有效并且更简洁。

主要区别在于，使用None作为密钥而不是custom（可能与列名冲突），apply_filters变得更强大。

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({None: qset}) # None is used as the key to specify that these are non-keyword filters

    return orm_filters

def apply_filters(self, request, applicable_filters):
    return self.get_object_list(request).filter(*applicable_filters.pop(None, []), **applicable_filters)
    # Taking the non-keyword filters out of applicable_filters (if any) and applying them as positional arguments to filter()

Django Tastypie高级过滤：如何使用Q对象进行复杂查找

3 个答案: