[英]XML of Django model and models related to it
I have a model class Equip, that it's a subclass of User (i think it doesn't matter) and a Player model. 我有一个模型类Equip,它是User(我认为没关系)和Player模型的子类。
class Team(auth.models.User):
emaile = models.EmailField('email',null=False,unique=True,)
class Player(models.Model):
name = models.CharField(max_length=50)
team = models.ForeignKey(Equip)
In views i have: 在意见上,我有:
class ConnegResponseMixin(TemplateResponseMixin):
def render_xml_object_response(self, objects, **kwargs):
xml_data = serializers.serialize(u"xml", objects, **kwargs)
return HttpResponse(xml_data, content_type=u"application/xml")
def render_to_response(self, context, **kwargs):
if 'extension' in self.kwargs:
try:
objects = [self.object]
except AttributeError:
objects = self.object_list
return self.render_xml_object_response(objects=objects)
else:
return super(ConnegResponseMixin, self).render_to_response(context
class teamDetail(DetailView, ConnegResponseMixin):
model = Equip
template_name = 'competition/team_detail.html'
def get_context_data(self, **kwargs):
context = super(teamDetail,self).get_context_data(**kwargs)
return context
How can I have a xml like: Team: Name:teamname email:teamemail players:{ player: name:playername player2: name:player2name 我该如何使用xml这样的XML:团队:名称:teamname电子邮件:teamemail玩家:{玩家:名称:playername玩家2:名称:player2name
I have tried to create a new class TeamXML 我试图创建一个新的类TeamXML
class TeamXML():
def __init__(self, equip):
self.username = equip.username
self.email = equip.correoe
self.isTeamValid = equip.isTeamValid
self.players = Jugador.objects.filter(team=equip)
and in the views I have changed the function render_xml_object_response to: 在视图中,我将函数render_xml_object_response更改为:
def render_xml_object_response(self, objects, **kwargs):
if objects[0].__class__ == Team:
objectA = TeamXML(objects[0])
xml_data = serializers.serialize(u"xml", objectA, **kwargs)
else:
xml_data = serializers.serialize(u"xml", objects, **kwargs)
return HttpResponse(xml_data, content_type=u"application/xml")
But it returns to me the error: 'TeamXML' object is not iterable. 但是它向我返回了错误:“ TeamXML”对象不可迭代。 How can I do TeamXML iterable or how can I do to return a XML with team and players like attribute of team.
如何做可迭代的TeamXML或如何返回具有球队和球员(例如球队属性)的XML。
Traceback:
File "/usr/local/lib/python2.7/dist-packages/django/core/handlers/base.py" in get_response
111. response = wrapped_callback(request, *callback_args, **callback_kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/views/generic/base.py" in view
69. return self.dispatch(request, *args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/views/generic/base.py" in dispatch
87. return handler(request, *args, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/views/generic/detail.py" in get
116. return self.render_to_response(context)
File "/home/eloi/Models/DjangoLoL/LoL/competition/views.py" in render_to_response
59. return self.render_xml_object_response(objects=objects)
File "/home/eloi/Models/DjangoLoL/LoL/competition/views.py" in render_xml_object_response
48. xml_data = serializers.serialize(u"xml", objectA, **kwargs)
File "/usr/local/lib/python2.7/dist-packages/django/core/serializers/__init__.py" in serialize
128. s.serialize(queryset, **options)
File "/usr/local/lib/python2.7/dist-packages/django/core/serializers/base.py" in serialize
52. for obj in queryset:
Exception Type: TypeError at /team/10.xml
Exception Value: 'TeamXML' object is not iterable
Thanks, 谢谢,
I finnally solved with dexml. 我终于用dexml解决了。 https://pypi.python.org/pypi/dexml/
https://pypi.python.org/pypi/dexml/
With this: Models.py 与此:Models.py
class PlayerXML(dexml.Model):
name = fields.String()
rol = fields.String()
email = fields.String()
def __init__(self, player):
self.name = player.name
self.rol = player.rol
self.email = player.email
class TeamXML(dexml.Model):
username = fields.String()
email = fields.String()
players = fields.List(Jugador)
def __init__(self, team):
self.username = team.username
self.email = team.correoe
for item in Jugador.objects.filter(team=team):
self.players.append(PlayerXML(item))
And Views.py: 和Views.py:
team_xml = TeamXML(objects[0])
team_xml = TeamXML(objects[0])
xml_data = team_xml.render()xml_data = team_xml.render()
return HttpResponse(xml_data, content_type=u"application/xml")
Thanks for the help :) 谢谢您的帮助 :)
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.