如何仅向graphene-django中的用户个人资料所有者显示特定字段?
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【中文标题】如何仅向graphene-django中的用户个人资料所有者显示特定字段?【英文标题】:How to show specific field only to the user profile owner in graphene-django? 【发布时间】:2020-02-07 21:11:12 【问题描述】:我的graphene-django
应用程序中有以下架构:
import graphene
from django.contrib.auth import get_user_model
from graphene_django import DjangoObjectType
class UserType(DjangoObjectType):
class Meta:
model = get_user_model()
fields = ("id", "username", "email")
class Query(object):
user = graphene.Field(UserType, user_id=graphene.Int())
def resolve_user(self, info, user_id):
user = get_user_model().objects.get(pk=user_id)
if info.context.user.id != user_id:
# If the query didn't access email field -> query is ok
# If the query tried to access email field -> raise an error
else:
# Logged in as the user we're querying -> let the query access all the fields
我希望能够通过以下方式查询架构:
# Logged in as user 1 => no errors, because we're allowed to see all fields
query
user (userId: 1)
id
username
email
# Not logged in as user 1 => no errors, because not trying to see email
query
user (userId: 1)
id
username
# Not logged in as user 1 => return error because accessing email
query
user (userId: 1)
id
username
email
我怎样才能做到只有登录用户才能看到自己个人资料的email
字段,而其他人不能看到其他人的电子邮件?
【问题讨论】:
看一下石墨烯中间件的authorization example。 @TomasLinhart 我已经经历过很多次了,但我不知道该怎么做。 你说得对,在这种情况下使用授权中间件可能有点多余。但是,解决方案取决于您是要返回没有email
字段的用户对象(由第一个代码块建议)还是引发错误(由第二个代码块建议)。如果是前者,只需在if info.context.user.id != user_id:
分支中执行user['email'] = None
或del user['email']
。如果是后者,请在此处引发异常。
@TomášLinhart 我稍微编辑了我的 graphql 和 python cmets。如果查询正在访问电子邮件并且未登录到用户,我只想引发错误。您介意将您的评论扩展到答案吗?
【参考方案1】:
这是我将根据 cmets 采取的方法。这里的主要问题是能够获取解析器中查询请求的字段列表。为此,我使用了改编自 here 的代码:
def get_requested_fields(info):
"""Get list of fields requested in a query."""
fragments = info.fragments
def iterate_field_names(prefix, field):
name = field.name.value
if isinstance(field, FragmentSpread):
results = []
new_prefix = prefix
sub_selection = fragments[name].selection_set.selections
else:
results = [prefix + name]
new_prefix = prefix + name + '.'
sub_selection = \
field.selection_set.selections if field.selection_set else []
for sub_field in sub_selection:
results += iterate_field_names(new_prefix, sub_field)
return results
results = iterate_field_names('', info.field_asts[0])
return results
其余的应该很简单:
import graphene
from django.contrib.auth import get_user_model
from graphene_django import DjangoObjectType
class AuthorizationError(Exception):
"""Authorization failed."""
class UserType(DjangoObjectType):
class Meta:
model = get_user_model()
fields = ("id", "username", "email")
class Query(object):
user = graphene.Field(UserType, user_id=graphene.Int())
def resolve_user(self, info, user_id):
user = get_user_model().objects.get(pk=user_id)
if info.context.user.id != user_id:
fields = get_requested_fields(info)
if 'user.email' in fields:
raise AuthorizationError('Not authorized to access user email')
return user
【讨论】:
【参考方案2】:目前的答案过于复杂。只需创建两个 ObjectType,例如:
class PublicUserType(DjangoObjectType):
class Meta:
model = get_user_model()
fields = ('id', 'username')
class PrivateUserType(DjangoObjectType):
class Meta:
model = get_user_model()
花了 4 个多小时尝试其他解决方案,然后才意识到原来如此简单
【讨论】:
【参考方案3】:我最后就是这样做的,查询自己的信息时返回email
的实际值,为别人返回None
:
import graphene
from django.contrib.auth import get_user_model
from graphene_django import DjangoObjectType
class UserType(DjangoObjectType):
class Meta:
model = get_user_model()
fields = ("id", "username", "email")
def resolve_email(self, info):
if info.context.user.is_authenticated and self.pk == info.context.user.pk:
return self.email
else:
return None
class Query(graphene.ObjectType):
user = graphene.Field(UserType, user_id=graphene.Int())
def resolve_user(self, info, user_id):
return get_user_model().objects.get(pk=user_id)
【讨论】:
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