Apollo客户端创建Project时GraphQLError：期望StatusInput为对象

问题详情 #

已在Apollo Sandbox中成功创建项目，返回结果如下：

{"data": {"createProject": {"_id": "65952c08576a8d317dc4df47", "description": "front end", "name": "new react", "status": [{"code": "a", "launch": "c", "content": "b", "planning": "d","qa": "f", "start": "g", "ux": "h" }]}}}

客户端侧定义了如下Mutation：

mutation createProject($name: String! $description: String! $status: [StatusInput]!) {    
  createProject(name: $name, description: $description, status: $status) { 
    _id 
    name 
    description 
    status {
      start 
      planning 
      ux 
      content 
      code 
      qa 
      launch 
    }
  }
}

提交处理函数代码：

const handleSubmit = async (event) => {
  event.preventDefault();

  try {
    const mutationRes = await createProject({
      variables: {
        name: formState.name,
        description: formState.description,
        status: formState.status,
      },
    });

    console.log(mutationRes);
  } catch (err) {
    console.log(err);
  }
};

遇到的问题 #

网络请求中能看到name、description和status参数，但接口返回400状态码，报错信息：

GraphQLError: Expected type "StatusInput" to be an object.

服务端TypeDef定义 #

• Input类型：

input StatusInput { 
  start: String! 
  planning: String! 
  ux: String! 
  content: String! 
  code: String! 
  qa: String! 
  launch: String! 
  project: ID! 
}

• Mutation定义：

createProject( name: String! description: String! status: [StatusInput]! ): Project

内容的提问来源于stack exchange，提问作者kiwizebon