graphql解析器中的模拟resolveInfo参数

Question

我创建了一个辅助函数，它将从解析器中提取请求的字段，

module.exports = {
  Query: {
    users: (_, data, ctx, resolveInfo) => {
      const requestedFields = parseRequestedFields(resolveInfo);
      return UsersService.findById(userId, requestedFields);
    }
  }
}

我想测试parseRequestedFields()一下我需要创建一个假的resolveInfo.

我如何resolveInfo在这里模拟对象？

更新

这是类型定义resolveInfo

type GraphQLResolveInfo = {
  fieldName: string,
  fieldNodes: Array<Field>,
  returnType: GraphQLOutputType,
  parentType: GraphQLCompositeType,
  schema: GraphQLSchema,
  fragments: { [fragmentName: string]: FragmentDefinition },
  rootValue: any,
  operation: OperationDefinition,
  variableValues: { [variableName: string]: any },
}