Python graphql 异常处理：是否期望得到具有 200 OK 响应的错误数组？

Question

根据https://www.howtographql.com/graphql-python/6-error-handling/ 中的文档，我使用raise GraphQLError 在我的 Flask GraphQL 应用程序变异函数中显示错误，如下所示：

import graphene
from graphql import GraphQLError

from ...extensions import db
from ...models import User as UserModel
from ..types import User as UserType

class Update(graphene.Mutation):
    class Input:
        id = graphene.ID(required=True)
        # phone = graphene.String()
        name = graphene.String(required=False, default_value=None)
        # active = graphene.Boolean()

    Output = UserType

    @staticmethod
    def mutate(root, info, **kwargs):
        user = graphene.Node.get_node_from_global_id(info, kwargs.pop('id'))
        # print(info.context)
        # if not user:
        raise GraphQLError('eeee')
        # user.update(**kwargs)
        # db.session.commit()

        return user

我期望得到类似 400 状态码的东西，带有 graphql 错误 json 模式。但我得到 200 并且异常打印在带有回溯的控制台中。我在这里做错了吗？

An error occurred while resolving field Mutation.updateUser
Traceback (most recent call last):
  File "/.local/share/virtualenvs/Server-CvYlbWSB/lib/python3.7/site-packages/graphql/execution/executor.py", line 447, in resolve_or_error
    return executor.execute(resolve_fn, source, info, **args)
  File "/.local/share/virtualenvs/Server-CvYlbWSB/lib/python3.7/site-packages/graphql/execution/executors/sync.py", line 16, in execute
    return fn(*args, **kwargs)
  File "/application/schema/mutation/user.py", line 40, in mutate
    raise GraphQLError('eeee')
graphql.error.base.GraphQLError: eeee
Traceback (most recent call last):
  File "/.local/share/virtualenvs/Server-CvYlbWSB/lib/python3.7/site-packages/graphql/execution/executor.py", line 447, in resolve_or_error
    return executor.execute(resolve_fn, source, info, **args)
  File "/.local/share/virtualenvs/Server-CvYlbWSB/lib/python3.7/site-packages/graphql/execution/executors/sync.py", line 16, in execute
    return fn(*args, **kwargs)
  File "/application/schema/mutation/user.py", line 40, in mutate
    raise GraphQLError('eeee')
graphql.error.located_error.GraphQLLocatedError: eeee

127.0.0.1 - - [17/Oct/2018 01:46:54] "POST /graphql? HTTP/1.1" 200 - 

Answer 1

显示堆栈跟踪似乎是有意的。您可以查看discussion on GitHub。以防万一链接失效，讨论的基础是graphql-core 库基本上会吃掉石墨烯抛出的所有错误并将它们放在results.errors 数组中，而不会将堆栈跟踪打印到sys.stderr。通常，这是不受欢迎的行为，因此它似乎在拉取请求中被更改。

---

如果您仍想模仿该行为，可以查看此 StackOverflow 答案以摆脱堆栈跟踪：You can turn off the traceback by limiting its depth。它应该仍然以这种方式显示在results.errors；但是请注意，这仍然会在控制台上打印错误消息，但不会打印堆栈跟踪。

如果您想完全摆脱控制台上的错误和堆栈跟踪（我不推荐这样做），您需要在应用程序中的某处捕获异常 > 突变解析器，以便错误仍然显示在results.errors 数组中。例如，您可以在 Flask 应用程序最初运行时执行此操作（尽管在这种情况下范围可能太大）。

try:
    app = Flask(__name__)
except GraphQLError as gqle:
    pass # ignore the error
except OtherErrorYouManuallyCall as oeymc:
    pass 
# Any other error will be thrown and show the stack trace