检索Django中通用外键关系的"多"结尾

Question

在Django中,当我请求具有多对多关系的资源时,我最终得到关系的子部分中的所有项目,即使那些与父项没有直接关系的项目也是如此.如果我向你展示代码(修剪的类只显示必要的内容)会更容易:

楷模

class Report(models.Model):
    name = models.CharField(max_length=255)
    slug = AutoSlugField(_('slug'), populate_from='name')
    wells = models.ManyToManyField(Well, null=True)
    uuid = UUIDField(editable=False, blank=True, version=4, unique=True)


class Well(models.Model):
    slug = AutoSlugField(_('slug'), populate_from='name')
    name = models.CharField(max_length=255)

class Node(models.Model):
    @property
    def well(self):
        raise NotImplementedError("The 'well' field must be implemented")
    //irrelevant GFK omitted
    page_content_type = models.ForeignKey(ContentType, null=True, blank=True, related_name='page')
    page_object_id = models.PositiveIntegerField(blank=True, null=True)
    page_content_object = generic.GenericForeignKey('page_content_type', 
     'page_object_id')

资源

class ReportResource(ModelResource):
    wells = fields.ManyToManyField(WellResource, 'wells', full=True)
    stock = fields.ForeignKey(TickerResource, 'stock', full=True)

    class Meta:
        queryset = Report.objects.all()
        resource_name = 'ticker_reports'

class WellResource(ModelResource):
    nodes = fields.ToManyField('wells.api.NodeResource', 'nodes', full=True)
    type = fields.ForeignKey(WellTypeResource, 'type', full=True)

    class Meta:
        queryset = Well.objects.all()
        resource_name = 'wells'

class NodeResource(ModelResource):
    order = fields.IntegerField()
    content_object = GenericForeignKeyField({
                                                Content: UUIDOnlyContentResource
                                            }, 'content_object', full=True)

    class Meta:
        queryset = Node.objects.all()
        resource_name = 'nodes'
        filtering = {
            'ticker_report': ALL_WITH_RELATIONS
        }

Ticker报告有很多Wells,这些Wells在所有Ticker报告中共享.不同的是你可以将节点绑定到Wells; 对于给定的股票报告,应显示的唯一节点是与该股票报告相关的节点.

因此,对于给定的股票报告和一组井,只应显示与该Ticker报告共享该GenericForeignKey的节点.

关系:

page_object_id,page_content_object,page_content_type是GenericForeignKey关系到报告

目前,显示所有节点(这是一个错误).

在TastyPie中,如何告诉它只显示相关对象而不是所有对象？

这是一个简短的python控制台,可以更加简洁地显示问题:

>>> r = Report.objects.get(id=1)                                                                                                        
>>> for well in r.wells.all():                                                                                                          
...     for node in well.nodes.all():                                                                                                   
...         print 'Node in Well {0} is {1}'.format(well, node)                                                                          
...                                                                                                                                     
Node in Well The Areas You Must Watch (the-areas-you-must-watch - Fancy List) is Apple Content #1:Apple (0)                             
Node in Well The Areas You Must Watch (the-areas-you-must-watch - Fancy List) is First Solar Content #1:first solar (0)                 
Node in Well Risks (risks - Headline and Lead) is Apple Content #2:Apple (0)                                                            
Node in Well Risks (risks - Headline and Lead) is First Solar Content #2:first solar (0)                                                
>>> 

SQL实际输出

SELECT node.id, node.uuid, node.order,node.content_type_id, node.object_id,
       node.page_content_type_id, node.page_object_id, node.well_id FROM node 
WHERE node.well_id = 1  
ORDER BY node.order ASC 

(修改后更容易阅读)

预期的SQL输出:

SELECT node.id, node.uuid, node.order,node.content_type_id, node.object_id,
       node.page_content_type_id, node.page_object_id, node.well_id FROM node 
WHERE node.well_id = 1  AND node.page_content_type_id = 99 /*Report Content TypeID */ AND node.page_content_object_id = 1 /*ReportID*/
ORDER BY node.order ASC 

预期产量:

Node in Well The Areas You Must Watch is Apple Content #1
Node in Well Risks is Apple Content #2:Apple (0)

如何过滤掉与Django和TastyPie的多对多关系的子端(尽管没有TastyPie这个问题很明显,让我相信这是一个结构性问题)

Answer 1

执行查询时

well.nodes.all()

这将通过您在模型中描述的关系获取与之相关的所有节点well.

听起来好像你想要通过通用外键关系将返回的节点限制为引用该Report对象的节点.是对的吗？如果是,则需要显式过滤节点,如下所示:rpage_content_object

r = Report.objects.get(id=1)
for well in r.wells.all():                                                                                                          
    for node in well.nodes.filter(page_object_id = r.id,
                                  page_content_type = ContentType.objects.get_for_model(r)):
        # ...

更新:我对TastyPie一无所知,我不得不说我真的不明白你想要做什么,但是在Django中,如果井和节点之间存在关系,那么你需要将它添加到你的模特.例如,如果每个Node属于一个Well,那么自然要做的就是向Node模型添加一个字段:

class Node(models.Model):
    # ...
    well = models.ForeignKey('Well')

然后,如果要查找属于特定Report对象的井的所有节点r,所述节点也r通过通用外键关系引用,则发出查询:

 Node.objects.filter(well__report = r,
                     page_object_id = r.id,
                     page_content_type = ContentType.objects.get_for_model(r))

如果你必须做很多事情,那么自然要做的就是在Report模型上添加一个方法.