如何将 json 列表字符串反序列化为对象字符串列表？

Question

我有一个字符串对象，我试图对其进行反序列化，但是当另一个 json 对象中有一个 json 对象时，它在反序列化时出现问题。

[
  {
    "attributes": {
      "type": "Account",
      "url": "/services/data/77.0/object/Account/1234"
    },
    "Id": "1234",
    "IsDeleted": false,
    "MasterRecordId": null,
    "Name": "Stevens Smith",
    "LastName": null,
    "FirstName": null,
    "Salutation": null,
  },
  {
    "attributes": {
      "type": "Account",
      "url": "/services/data/v77.0/object/Account/12345"
    },
    "Id": "12345",
    "IsDeleted": false,
    "MasterRecordId": null,
    "Name": "Mr John Smith",
    "LastName": null,
    "FirstName": null,
    "Salutation": null,
  }
]

我的班级定义如下：

    public Account() {
    public string? attributes;
    public string? id;
    public string? isDeleted;
    public string? masterRecordId;
    public string? name;
    public string? lastName;
    public string? firstName;
    public string? salutation;
}

我收到的错误消息是解析值时遇到意外字符：{。 路径“[0].attributes”，第 3 行，位置 19。

Answer 1

你应该有两个类

public class Account() {
    public Attrubute attributes;
    public string? id;
    public string? isDeleted;
    public string? masterRecordId;
    public string? name;
    public string? lastName;
    public string? firstName;
    public string? salutation;
}

public class Attrubute() {
    public string? type;
    public string? url;
}

如果您只想将Attrubute对象保存为字符串，您可以这样做（语法可能会有所不同，具体取决于您是否使用 newtonsoft）

public class Account() {
    public Attrubute attributes;
    public string? id;
    public string? isDeleted;
    public string? masterRecordId;
    public string? name;
    public string? lastName;
    public string? firstName;
    public string? salutation;
    public string attributesString => attributes != null ? JsonConvert.Serialize(attributes) : string.empty;
}

你不想定义Attrubute类。 你可以使它成为动态对象，如果你想要字符串版本 - 你可以在你的类中进行转换。 所以像这样

public class Account {
    public string? id;
    public string? isDeleted;
    public string? masterRecordId;
    public string? name;
    public string? lastName;
    public string? firstName;
    public string? salutation;
    
    public dynamic attributes;
    public string Attributes => attributes != null ? JsonConvert.SerializeObject(attributes) : string.Empty;
}

Answer 2

你的很多变量都是错误的类型。 使用正确的类型要安全得多——这样，如果你重新序列化为 JSON，它将保持有效（例如 isDeleted——即使你将它解析为字符串，重新序列化会将其存储为：“false”而不是：false） .

public class Account
{
    public Attrubute attributes;
    public string stringAttributes //if you need a string accessor to attributes you can just use a property
        {
            get {
                return $"{attributes?.type}:{attributes?.url}";
            }
        }
    public string? id;
    public bool isDeleted;
    public string? masterRecordId;
    public string? name;
    public string? lastName;
    public string? firstName;
    public string? salutation;
}

public class Attrubute
{
    public string? type;
    public string? url;
}

如果您确实想将其保留为字符串，更好的选择是将其保留为 json 而不是 JObject，并在需要时调用.ToString() 返回 json 值

public class Account
{
    public JObject attributes;
    public string? id;
    public JObject isDeleted;
    public string? masterRecordId;
    public string? name;
    public string? lastName;
    public string? firstName;
    public string? salutation;
    public string GetAttributesJson() {
         return attributes.ToString();
    }
}

Answer 3

如果你需要属性作为字符串，你可以使用一个 counstructor 来实现，但是在你必须通过添加 getter/setter 来修复类之前

List<Account> accounts = JsonConvert.DeserializeObject<List<Account>>(json);

public class Account
{
    public int? id { get; set; }
    public string? attributes { get; set; }
    //.... another properties
    [JsonConstructor]
    public Account(JObject attributes)
    {
        this.attributes = string.Join(",", attributes.Properties());

       //or a json string
       this.attributes =  attributes.ToString();

      // or a url
       this.attributes = (string)attributes["url"]

    }
}