Question

我已经对这个问题进行了大量的搜索，并在Stack Overflow上经历了一系列相关的问题，但似乎并不是关于如何实现多对多关系的明确答案。通过＆＃34;中间模型（或者我错过了它）。

我有一个名为Sample的模型，它与Region有多对多的关系。有一个连接两者的中间模型，名为SampleRegion。我目前没有保存关于中间模型的任何额外信息，但我可能在将来。

以下是我的模特：

class Sample(models.Model):
    sample_id = models.BigIntegerField(primary_key=True)
    description = models.TextField(blank=True)
    objects = models.GeoManager()
    regions = ManyToManyField(Region, through='SampleRegion')
    class Meta:
        db_table = u'samples'
    def save(self, **kwargs):
        # Assign a sample ID only for create requests
        if self.sample_id is None:
            try: id = Sample.objects.latest('sample_id').sample_id + 1
            except Sample.DoesNotExist: id = 1
            self.sample_id = id
        super(Sample, self).save

class Region(models.Model):
    name = models.CharField(max_length=100, unique=True)
    def __unicode__(self):
        return self.name
    class Meta:
        db_table = u'regions'

class SampleRegion(models.Model):
    sample = models.ForeignKey('Sample')
    region = models.ForeignKey(Region)
    class Meta:
        unique_together = (('sample', 'region'),)
        db_table = u'sample_regions'

这是我用来编写资源的一种方法。它不正确，我无法找到正确的方法：

class SampleResource(ModelResource):
    regions = fields.ToManyField("tastyapi.resources.RegionResource",
                                  "regions")
    class Meta:
        queryset = models.Sample.objects.all()
        allowed_methods = ['get', 'post', 'put', 'delete']
        authentication = ApiKeyAuthentication()
        authorization = ObjectAuthorization('tastyapi', 'sample')
        excludes = ['user', 'collector']
        filtering = {
                'version': ALL,
                'sesar_number': ALL
                }
        validation = VersionValidation(queryset, 'sample_id')

    def hydrate_regions(self, bundle): 
        # code to create a new SampleRegion object by getting a list of 
        # regions from bundle.data['regions']

class RegionResource(ModelResource):
    class Meta:
        queryset = models.Region.objects.all()
        allowed_methods = ['get']
        resource_name = "region"
        filtering = {
                'region': ALL,
                }

这是我发布POST请求的方式：

post_data = {
    'regions': ["/tastyapi/v1/region/2/"],
    'description': 'Created by a test case',
}

client.post('/tastyapi/v1/sample/', data = post_data,
            authentication = credentials, format = 'json')

此请求不起作用，因为到目前为止，bundle.data [＆＃39;地区＆＃39;]为None 它到达hydrate_regions。

有没有人对如何实施这种情况有任何建议？

Answer 1

我几天前就想到了这一点。这是我发现的......

如果您没有明确创建中间表，Django会为您创建M2M关系。但是，如果您明确使用中间表，那么您负责在中间表中创建记录。为了在Tastypie中工作，我不得不重写save_m2m方法，在中间表中显式创建一条记录，将我刚刚创建的样本和现有区域链接起来。

这就是我resources.py的相关部分现在的样子：

class SampleResource(ModelResource):
    regions = fields.ToManyField("tastyapi.resources.RegionResource",
                                 "regions")

    class Meta:
        queryset = models.Sample.objects.all()
        allowed_methods = ['get', 'post', 'put', 'delete']
        authentication = ApiKeyAuthentication()
        authorization = ObjectAuthorization('tastyapi', 'sample')
        excludes = ['user', 'collector']
        filtering = {
                'regions': ALL_WITH_RELATIONS,
                }
        validation = VersionValidation(queryset, 'sample_id')

    def save_m2m(self, bundle):
        for field_name, field_object in self.fields.items():
            if not getattr(field_object, 'is_m2m', False):
                continue

            if not field_object.attribute:
                continue

            for field in bundle.data[field_name]:
                kwargs = {'sample': models.Sample.objects.get(pk=bundle.obj.sample_id),
                          'region': field.obj}

                try: SampleRegion.objects.get_or_create(**kwargs)
                except IntegrityError: continue

class RegionResource(BaseResource):
    class Meta:
        queryset = models.Region.objects.all()
        authentication = ApiKeyAuthentication()
        allowed_methods = ['get']
        resource_name = "region"
        filtering = { 'region': ALL }

Django Tastypie：通过＆＃34;通过＆＃34;关系

1 个答案: