Django 和 Tastypie 的相关资源

Question

编辑 2： 我取得了一些进展并更新了我的代码如下：

我在models.py中有这三个模型：

 class Variables(models.Model):
    variableid = models.IntegerField(db_column='VariableID', primary_key=True)
    variableName = models.CharField(db_column='VariableNameCV', max_length=255) 

    class Meta:
        managed = False
        db_table = 'variables'

class Results(models.Model):
    resultid = models.BigIntegerField(db_column='ResultID', primary_key=True) 
    variable = models.ForeignKey('Variables', related_name='results', db_column='VariableID')
    units = models.ForeignKey('Units', related_name='units', db_column='UnitsID')

    class Meta:
        managed = False
        db_table = 'results'

class Units(models.Model):
    unitsid = models.IntegerField(db_column='UnitsID', primary_key=True)
    unitstype = models.CharField(db_column='UnitsTypeCV', max_length=255)
    unitsname = models.CharField(db_column='UnitsName', max_length=255)

    class Meta:
        managed = False
        db_table = 'units'

我定义了这些 tastypie 模型资源：

class UnitsResource(ModelResource):
    class Meta:
        queryset = Units.objects.all()
        resource_name = 'units'
        collection_name = 'units'
        authorization = Authorization()
        serializer = Serializer()


class ResultsResource(ModelResource):
    units = fields.ForeignKey(UnitsResource, attribute='units', full=True, null=True);
    class Meta:
        queryset = Results.objects.all() 
        resource_name = 'results'
        collection_name = 'results'
        authorization = Authorization()
        serializer = Serializer()

class VariablesResource(ModelResource):
    results = fields.ToManyField(ResultsResource, attribute='results', related_name='results', full=True, null=True);
    class Meta:
        queryset = Variables.objects.all()
        resource_name = 'variables'
        collection_name = 'variables'
        authorization = Authorization()
        serializer = Serializer()

现在输出是这样的：

<variable>
 <variableName>...</variableName>
 <variableid>...</variableid>
 <result type="list">
  <object>
    <resultid>...</resultid>
    <units>
      <unitname>...</unitname>
      <unittype>...</unittype>
    </units>
  </object>
</variable>

当我请求此 URL (http://127.0.0.1:8000/api/v1/variables/4/) 时，我试图获得以下输出：

<variable>
 <variableName>...</variableName>
 <variableID>...<variableID>
 <unit>
   <unitName>...</unitName>
   <unitType>...</unitType>
 </unit>
</variable>

看来我必须经过 ResultsResource 才能到达 UnitsResource，但是我怎样才能让我的 XML 输出忽略来自 ResultsResource 的东西？作为附带问题，我可以查看哪些文档来帮助我解决这个问题？

Answer 1

我想我已经弄明白我需要什么了。这产生了我想要的结果：

class VariablesResource(ModelResource):
    results = fields.ToManyField(ResultsResource, attribute='results', related_name='results', full=True, null=True);
    class Meta:
        queryset = Variables.objects.all() 
        resource_name = 'variables'
        collection_name = 'variables'
        authorization = Authorization()
        serializer = Serializer()

    def dehydrate(self, bundle):
        variablesBundle = {}
        unitsBundle = {}
        for k, v in bundle.data.iteritems():
            if (k != u'results'):
                variablesBundle[k] = v 
            else:
                for i in v:        
                    unitsBundle = i.data['units'] 
        variablesBundle['unit'] = unitsBundle
        return variablesBundle

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