无法使用JsonApiFramework获取相关资源

Question

使用https://github.com/scott-mcdonald/JsonApiFramework框架

我从UI中发送以下修补程序：

{"data":{"id":"4221fe2f-ca7b-425d-b45a-a09f4936a22b","attributes":{"contact-trade-name":"asdad","contact-name":"aasd","contact-phone":"asdad","contact-email":"asdad@adasd.asd","created-date":"2016-12-19T14:07:59.833","status":"AwaitingBillData"},"relationships":{"share-user":{"data":{"type":"users","id":"5078b2ee-2348-44f5-95a0-3c14c2ad2b1d"}}},"type":"jobs"}}

在我的服务器上，我执行以下操作：

using (var documentContext = new DocumentContext(HttpContext.Current.Request.Url, document)) { var job = documentContext.GetResource<Job>(); var relationships = documentContext.GetResourceRelationships(job); var shareUserRel = relationships["share-user"]; var shareUser = documentContext.GetRelatedResource<Data.ServiceModel.User>(shareUserRel); }

然而，shareUser总是null

当我检查shareUserRel时，它有以下几点：

{ResourceIdentifier [type=users id=5078b2ee-2348-44f5-95a0-3c14c2ad2b1d]}

如果有帮助的话。

有什么明显的证据表明我做错了什么吗？

Answer 1

要将json:api文档格式化为可读性，如下所示：

{
  "data": {
    "type": "jobs"
    "id": "4221fe2f-ca7b-425d-b45a-a09f4936a22b",
    "attributes": {
      "contact-trade-name": "asdad",
      "contact-name": "aasd",
      "contact-phone": "asdad",
      "contact-email": "asdad@adasd.asd",
      "created-date": "2016-12-19T14:07:59.833",
      "status": "AwaitingBillData"
    },
    "relationships": {
      "share-user": {
        "data": {
          "type": "users",
          "id": "5078b2ee-2348-44f5-95a0-3c14c2ad2b1d"
        }
      }
    }
  }
}

GetRelatedResource返回null的原因是，您要求文档上下文返回文档中包含的实际用户资源，文档中不存在该资源。

当客户端在补丁中发送这种类型的json:api文档时，您将指示服务器用给定的属性更新作业资源，并将此作业资源与在to-one资源链接中指定的现有用户资源关联起来。

在示例中有了relationship对象之后，更正确的内容如下：

if (shareUserRel.IsResourceLinkageNullOrEmpty())
{
  // Clear any linkage between the job resource and any current user
  ...
}
else
{
  var shareUserResourceIdentifier = shareUserRel.GetToOneResourceLinkage();
  // Create linkage between the job resource and existing user identified
  // by shareUserResourceIdentifier
  ...
}

现在你如何处理“.”取决于您的数据层。