Question

我的模型中定义为models.ForeignKey的字段未显示在Tastypie中。除客户端字段外，显示所有字段。快速而肮脏的修复是在Statement模型中添加另一个字段，如

ClientID = models.IntegerField(db_column='Client_id', max_length=32)

但这对我来说似乎不对。有谁知道更好的解决方案？

models.py

class Client(models.Model):
    ID = models.AutoField(primary_key=True)
    F1 = models.CharField(max_length=256, null=True)
    F2 = models.CharField(max_length=256, null=True)


class Statement(models.Model):
    ID = models.AutoField(primary_key=True)
    Client = models.ForeignKey(Client, related_name='statements')
    State = models.CharField(max_length=256, null=True)
    Address = models.CharField(max_length=256, null=True)

api.py

class StatementResource(ModelResource):
    class Meta:
        queryset = Statement.objects.all()
        resource_name = 'client'
        allowed_methods = ['get']
        include_resource_uri = False

class ClientResource(ModelResource):
    statements = fields.ToManyField(StatementResource, 'statements', null=True, blank=True, full=True)
    class Meta:
        queryset = Client.objects.all()
        resource_name = 'client'
        allowed_methods = ['get']
        include_resource_uri = False

Answer 1

在Tastypie中，您必须在ModelResources定义中定义所有关系字段。因此，如果您在模型中有一些说外键，则必须在模型资源中再次定义该外键。原因是Tastypie必须知道如何处理它。必须知道它的序列化，授权，认证等。该领域被忽视的其他方式。你必须传递那个相关的资源，就是这样。

class StatementResource(ModelResource):
    Client = fields.ForeignKey(ClientResource, 'Client')
    class Meta:
        queryset = Statement.objects.all()
        resource_name = 'client'
        allowed_methods = ['get']
        include_resource_uri = False

ForeignKey未在Tastypie中显示

1 个答案: