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python - 可扩展项目的 graphql 代码组织

转载 作者:行者123 更新时间:2023-12-05 07:29:24 36 4
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假设我们有一个包含大量应用程序的大项目,这会导致大量查询和变更。对于这样的项目,你们如何处理 graphql 代码架构。让我们举个例子。

我有一个名为帐户的应用程序。它将具有与用户和配置文件相关的查询和突变。我使用的文件夹结构是每个应用程序都会有 graphql 文件夹,然后会有 schema.py 和 mutations.py。代码暂时是这样安排的

架构.py

class User(DjangoObjectType):

class Meta:
model = CustomUser
filter_fields = {
'email': ['exact', ],
}
exclude_fields = ('password', 'is_superuser', )
interfaces = (relay.Node, )


class ProfileNode(DjangoObjectType):

class Meta:
model = Profile
interfaces = (relay.Node, )


class UserQuery(object):

user = relay.Node.Field(User)
users = DjangoFilterConnectionField(User) # resolve_users is not needed now


class ProfileQuery(object):

profile = relay.Node.Field(ProfileNode)
profiles = DjangoFilterConnectionField(ProfileNode)


class UserProfile(ObjectType):

profile = Field(ProfileNode)

def resolve_profile(self, info, **kwargs):
if id is not None and info.context.user.is_authenticated:
profile = Profile.objects.get(user=info.context.user)
return profile
return None


class Viewer(ObjectType):

user = Field(User)

def resolve_user(self, info, **kwargs):
if info.context.user.is_authenticated:
return info.context.user
return None

突变.py

class Register(graphene.Mutation):
"""
Mutation to register a user
"""
class Arguments:

first_name = graphene.String(required=True)
last_name = graphene.String(required=True)
email = graphene.String(required=True)
password = graphene.String(required=True)
password_repeat = graphene.String(required=True)

success = graphene.Boolean()
errors = graphene.List(graphene.String)

def mutate(self, info, first_name, last_name, email, password, password_repeat):
# console.log('info', info, first_name, last_name, email, password)
if password == password_repeat:
try:
user = CustomUser.objects.create(
first_name=first_name,
last_name=last_name,
email=email,
is_active=False
)
print ('user', user)
user.set_password(password)
user.save()
if djoser_settings.get('SEND_ACTIVATION_EMAIL'):
send_activation_email(user, info.context)
return Register(success=bool(user.id))
# TODO: specify exception
except Exception:
errors = ["email", "Email already registered."]
return Register(success=False, errors=errors)
errors = ["password", "Passwords don't match."]
return Register(success=False, errors=errors)

根架构

// just to show the number of mutations just for account apps.
from accounts.graphql.mutations import (
Activate,
DeleteAccount,
Login,
RefreshToken,
Register,
ResetPassword,
ResetPasswordConfirm,
)
from accounts.graphql.schema import Viewer, UserProfile


class Mutation(company_mutation.Mutation, graphene.ObjectType):

activate = Activate.Field()
debug = graphene.Field(DjangoDebug, name='__debug')


class Query(company_schema.Query, graphene.ObjectType):
viewer = graphene.Field(Viewer)
user_profile = graphene.Field(UserProfile)
debug = graphene.Field(DjangoDebug, name='__debug')

@staticmethod
def resolve_viewer(self, info, **kwargs):
if info.context.user.is_authenticated:
return info.context.user
return None

@staticmethod
def resolve_user_profile(self, info, **kwargs):
if info.context.user.is_authenticated and id:
return info.context.user
return None


schema = graphene.Schema(query=Query, mutation=Mutation)

你可以看到帐户应用程序的变化。有很多突变,考虑到所有应用程序时还会有更多突变。你们是如何处理这样的事情的?

最佳答案

我认为仅基于schema.pyqueries.pymutations.py等文件的项目组织对于大型项目来说是非常糟糕的.

这就像用 models.pyfields.pyutils.py 等文件组织您的模型部件 ...

特别是,在您的 queries.py 文件某处定义的 GraphQL ObjectType 可以返回或用作突变的输入。

所以我更喜欢基于对象及其逻辑关系的结构

schema/
__init__.py : contains your root schema actual code
viewer.py : contains the ObjectType Viewer
user/
user.py : contains the ObjectType User and UserQuery
profile.py : contains the ObjectType Profil
account : contains the account management mutations
login.py : contains the login / logout mutations
...

注意:您可以像查询一样嵌入突变,以便能够获得突变查询,例如:

mutation {
account {
delete(id: 12) {
status
}
}
}

关于python - 可扩展项目的 graphql 代码组织,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52809585/

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