Question

我正在将graphene-django用于api。我正在尝试创建一个具有公司外键的品牌突变。当我突变时，出现以下错误“'input'是print（）的无效关键字参数”。我不知道为什么会引发此错误。

这是我的突变

class BrandInput(graphene.InputObjectType):

    company = graphene.List(CompanyInput)
    name = graphene.String()
    website = graphene.String()
    description = graphene.String()
    country = graphene.String()
    city = graphene.String()
    address = graphene.String()


    class CreateBrand(graphene.Mutation):

        class Arguments:

            input = BrandInput(description="These fields are required", required=True)


        class Meta:

            description = "Create Brand Mutation"

        errors = graphene.String()
        brand = graphene.Field(BrandNode)

        @staticmethod
        def mutate(root, info, **args):
            print('args', args, **args)
            if not info.context.user.is_authenticated:
                return CreateBrand(errors=json.dumps('Please Login to list your brand'))
            try:
                company = models.Company.objects.get(slug=args.get('input')['company'])
                if company:
                    brand = models.Brand.objects.create(
                        company=company,
                        name=args.get('input')['name'],
                        slug = args.get('input')['slug'],
                        website = args.get('input')['website'],
                        description = args.get('input')['description'],
                        country = args.get('input')['country'],
                        city = args.get('input')['city'],
                        address = args.get('input')['address'],
                    )
                    return CreateBrand(brand=brand, errors=null)
            except models.Company.DoesNotExist:
                return CreateBrand(errors=json.dumps('Company should be required'))

我对company = graphene.List(CompanyInput)感到怀疑，因此我将其更改为company = graphene.String()并提供了公司的子弹，以便在更改品牌时可以找到公司实例。但是我遇到了同样的错误。

突变查询是

mutation {
  createBrand(input: {company: "wafty-company", name: "Wafty Brand", website: "www.wafty.com", description: "Wafty brand description", country: "Nepal", city: "Kathmandu", address: "Baneshwor", pinCode: "44600", operationDate: "2018-10-02 15:32:37", franchisingDate: "2018-10-02 15:32:37", numberOfFranchises: "0-10", numberOfOutlets: "0-10"}) {
    errors
    brand {
      name
      slug
      website
    }
  }
}

Answer 1

当您尝试将类似**args的参数传递给print()时，此参数将被解压缩为关键字参数，这将引发错误，因为print()不希望有此类参数就像mutate()方法一样。因此，您需要删除**args：

print('args', args)

Answer 2

正如其他人已经指出的那样，不清楚您正在尝试使用

实现什么

print('args', args, **args)

行。但是，无论如何，python的print函数的语法为（ref）：

print(object(s), separator=separator, end=end, file=file, flush=flush)

并赋予**<variableName>会使函数混淆，从而使错误消失。如果您有kwargs的字典，并且要打印值，则可以使用*kwargs.values()的语法。例如：

args = [1, 2]
kwargs = {'a':3, 'b':4}

print(*args, *kwargs.values())

将打印

1 2 3 4

和其他组合是：

print(kwargs)           # --> {'a': 3, 'b': 4}
print(*kwargs)          # --> a f
print(kwargs.values())  # --> dict_values([3, 4])
print(*kwargs.values()) # --> 3 4

this post中此处更相关的讨论。

“输入”是print（）的无效关键字参数

2 个答案: