C#中使用JsonConvert解析JSON

1.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);

2.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)

  

原文地址:https://www.cnblogs.com/wfy680/p/12664401.html