如何仅向graphene-django中的用户个人资料所有者显示特定字段？答案

【问题标题】：How to show specific field only to the user profile owner in graphene-django?如何仅向graphene-django中的用户个人资料所有者显示特定字段？
【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？

标签： python django graphql graphene-django

【解决方案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】：

【解决方案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】：

【解决方案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】：