如何将 Json 字符串反序列化为自定义 DTO C# 对象？

Question

我有 DTO 类。我收到了这样的 Json 序列化字符串，我该如何分配到上述 DTO 中？ 我正在尝试反序列化 json 字符串，因为它对我不起作用。

RequestMetaDataXml MetaDataDTO =new RequestMetaDataXml();
                MetaDataDTO = JsonConvert.DeserializeObject<RequestMetaDataXml>(arguments["MetaDataDTO"]);

我用下面的方法生成Json

var MetaDataDTO = {
        Title:replacedtitle ,
        Expirydate: expirydate,
        AllowDownload: checkallowdownload,
        IsShare: chkAllowShare,
        IncludeMetadata: chkincludeMetadata,
        IsReel: "false",
        IsSecuredPublish: IsSecuredPublish,
        Notifications: NotificationId,
        CoverArt: { UploadedFileName: UploadedFileName },
        ProfileInfo: {
            WaterMark: {
                VideoWatermark: {
                    WaterMarkInfo: {
                        Type : WaterMarkInfoType,
                        FreeText : WaterMarkText,
                        Position: WaterMarkPosition,
                        Size: WaterMarkSize,
                        LogoId : LogoId
                    }
                }
            }
        }
    }

如何在 c# 代码中反序列化 Json 的值以及如何将这些值分配到我的自定义 DTO 中？

Answer 1

让生活更轻松，并使用库为您进行反序列化。我常用的一个推荐和我使用的是 Newtonsoft.Json，反序列化就像这样简单：

JsonConvert.DeserializeObject<RequestMetaDataXml>(jsonStringReceived);

Answer 2

您必须修改一件事，json 或 DTO 类，因为 json 有一个用于 WaterMark、VideoWatermark 等的对象，而 DTO 类说这些是 List 类型的对象。此外，您还需要一个包装类，因为 RequestMetaDataXml 被包装在另一个对象中。

包装类：

[DataContract]
public class Wrapper
{
    [DataMember(Name = "RequestMetaDataXml")]
    public RequestMetaDataXml RequestMetaDataXml { get; set; }
}

更改后您的其他课程是：

[DataContract]
public class RequestMetaDataXml
{
    [DataMember(Name = "Title")]
    public string Title { get; set; }

    [DataMember(Name = "ExpiryDate")]
    public string ExpiryDate { get; set; }

    [DataMember(Name = "AllowDownload")]
    public string AllowDownload { get; set; }

    [DataMember(Name = "IsShare")]
    public string IsShare { get; set; }

    [DataMember(Name = "IncludeMetadata")]
    public string IncludeMetadata { get; set; }

    [DataMember(Name = "IsReel")]
    public string IsReel { get; set; }

    [DataMember(Name = "IsSecuredPublish")]
    public string IsSecuredPublish { get; set; }

    [DataMember(Name = "Notifications")]
    public NotificationId Notifications { get; set; }

    [DataMember(Name = "CoverArt")]
    public CoverArt Coverart { get; set; }

    [DataMember(Name = "ProfileInfo")]
    public ProfileInfo Profileinfo { get; set; }
}


[DataContract]
public class NotificationId
{
    [DataMember(Name = "Id")]
    public string Id { get; set; }

    [DataMember(Name = "Type")]
    public string Type { get; set; }
}

[DataContract]
public class CoverArt
{
    [DataMember(Name = "UploadedFileName")]
    public string UploadedFileName { get; set; }
}

[DataContract]
public class ProfileInfo
{
    [DataMember(Name = "Watermark")]
    public WaterMark WaterMark { get; set; }
}

[DataContract]
public class WaterMark
{
    [DataMember(Name = "VideoWatermark")]
    public VideoWatermark VideoWatermark { get; set; }
}

[DataContract]
public class VideoWatermark
{
    [DataMember(Name = "WaterMarkInfo")]
    public WaterMarkInfo WaterMarkInfo { get; set; }
}

[DataContract]
public class WaterMarkInfo
{
    [DataMember(Name = "Type")]
    public string Type { get; set; }

    [DataMember(Name = "FreeText")]
    public string FreeText { get; set; }

    [DataMember(Name = "Position")]
    public string Position { get; set; }

    [DataMember(Name = "Size")]
    public string Size { get; set; }

    [DataMember(Name = "LogoId")]
    public string LogoId { get; set; }
}

您必须添加Newtonsoft.Json 的引用，然后使用

 var wrapper = JsonConvert.DeserializeObject<Wrapper>(json);
 RequestMetaDataXml RequestMetaDataXml = wrapper.RequestMetaDataXml;

如果您不想将List 更改为单个对象，则必须更新您的json 并将List 类型对象的值放在[ ] 大括号中 例如

"Notifications": [{
        "Id": "checktest1@ssss.com",
        "Type": "ExtUser"
 }],

等等。

但无论如何你都需要Wrapper 类。

Answer 3

        var ser = new XmlSerializer(typeof(RequestMetaDataXml));
        var settings = new XmlSerializerNamespaces();
        settings.Add("", "");
        using (var sw = new StringWriter())
        {
            ser.Serialize(sw, o.RequestMetaDataXml, settings);
            var t = sw.ToString();
        }