C#中使用JsonConvert解析JSON_c# jsonconvert.serializeobject

1、首先下载 Newtonsoft.Json.dll 文件，然后添加Newtonsoft.Json的引用

2、using Newtonsoft.Json

3、JSON序列化

string JsonStr= JsonConvert.SerializeObject(Entity);
 
public class RecordResult
{
[JsonProperty("status")]
public int Status { get; set; }
[JsonProperty("error")]
public int Error { get; set; }
}
 
RecordResult result = new RecordResult();
 
result.Status = 1;
 
result.Error = "Error message";
 
string jsonStr = JsonConvert.SerializeObject(result);

4、JSON反序列化

Class model = JsonConvert.DeserializeObject<Class>(jsonstr);
 
var result = (RecordResult)JsonConvert.DeserializeObject(jsonstring, typeof(RecordResult))
 
//或者
 
var result = JsonConvert.DeserializeObject<RecordResult>(jsonstring)

原文链接：https://www.cnblogs.com/kevin860/p/11113112.html