C# 将属性附加到c中序列化的json字符串#

我有一个要序列化的类

public class DesignEntity
{
    public string DesignName { get; set; }
    public string DesignID { get; set; }
    [ScriptIgnore]
    public string designDetails { get; set; }
}

属性

designDetails

已包含序列化数据，因此无需再次序列化

DesignEntity designSessionEntity = null;
JavaScriptSerializer jSerializer = null;
string designJsonSession = "null";

designSessionEntity = new DesignEntity();
jSerializer = new JavaScriptSerializer();

designSessionEntity = /* */ // Get Values via some method
designJsonSession = jSerializer.Serialize(designSessionEntity); 
// Now append designDetails to resultant string json

---

序列化完成后，我想将

designDetails

附加到

designJsonSession

。如何实现相同的功能？

首先，在

设计实体

类中再添加一个字段。其类型是从中序列化的

designDetails

json的合同类

public class DesignEntity
{
    public string DesignName { get; set; }
    public string DesignID { get; set; }
    /// new field
    public DesignDetailsObject DesignDetails { get; set; }
    [ScriptIgnore]
    public string designDetails { get; set; }
}

然后，像往常一样获取对象，首先将json反序列化到新字段中，然后将其序列化

var designSessionEntity = new DesignEntity();
var jSerializer = new JavaScriptSerializer();

designSessionEntity = /// Get Values via some method

/// assign the design details with deserialized object
designSessionEntity.DesignDetails = jSerializer.Deserialize<DesignDetailsObject>(designSessionEntity.designDetails);

/// serialize it again
var designJsonSession = jSerializer.Serialize(designSessionEntity); 

用法：

var fieldAppendedJson = JsonAppender(json, new List<string> { "designDetails " }, designSessionEntity.designDetails);

var fieldAppendedJson=JsonAppender（json，新列表{“designDetails”}，designSessionEntity.designDetails）；

反序列化、添加并再次序列化。显示初始格式和所需结果。不清楚您想要什么输出格式。您为什么这样做：

string designJsonSession=“null”？您只需执行
string designJsonSession=null
@Sybren实际上不需要分配
null
-这是默认值。此外，没有必要将声明和分配分开：）@YeldarKurmangaliyev我知道没有必要这样做。但是看看他的代码，他想分配
null
。什么是
DesignDetailsObject
？这应该是json
designDetails
首先从中序列化的类。
var fieldAppendedJson = JsonAppender(json, new List<string> { "designDetails " }, designSessionEntity.designDetails);