I am using below code to Deserialize Object. objectJSonRequest2 = Newtonsoft.Json.JsonConvert.DeserializeObject<dynamic>(JSonRequest1);
DateFormatString="yyyy-MM-dd HH:mm:ss"};varjsonObject = JsonConvert.DeserializeObject<dynamic>(jsonString, jSetting);//dynamic动态解析varapiResourcesStr ="";if(((Newtonsoft.Json.Linq.JContainer)jsonObject).HasValues &&((Newtonsoft.Json.Linq.JContainer)jsonObject.ApiResources).HasValues) apiResou...
};varsr =newStringReader(jsonText);varreader =newJsonTextReader(sr);varresult = (T)json.Deserialize(reader,typeof(T)); reader.Close();returnresult; } } } 解决方法 在调用通过json反序列化的dynamic对象时,要先强制转换为对应的类型 代码: if(ContainChinese((jsonData.FY_Subtitle_CN ??"").ToS...
As per the Newtonsoft Documentation you can also deserialize to an anonymous object like this: var definition = new { Name = "" }; string json1 = @"{'Name':'James'}"; var customer1 = JsonConvert.DeserializeAnonymousType(json1, definition); Console.WriteLine(customer1.Name); // James ...
JSON The JSON string below is a simple response from an HTTP API call, and it defines two properties:IdandName. {"Id": 1, "Name": "biofractal"} 1. C# UseJsonConvert.DeserializeObject<dynamic>()to deserialize this string into a dynamic type then simply access its properties in the usual...
C#Newtonsoft.Json反序列化为dynamic对象之后的使⽤通过Newtonsoft.Json将⼀个json类型的字符串反序列化为dynamic后直接使⽤报错 源代码：namespace ConsoleApplication1 { class Program { static void Main(){ var data = "{\"C_Describe\":\"测试\",\"FY_Subtitle\":\"测试\",\"MAX_ZJ\":20000,\"...
dynamic obj = JsonConvert.DeserializeObject<ExpandoObject>(jsonString); 如果你想将JSON数据转换为匿名类型对象,可以使用var关键字。例如: 代码语言:txt 复制 var obj = JsonConvert.DeserializeObject(jsonString); 需要注意的是,使用Newtonsoft解析JSON时,你需要确保JSON数据的格式是有效的,否则可能会抛出异常。 Newton...
C# Json隐式Newtonsoft DeserializeObject方法的应用场景包括但不限于: 前后端数据交互:在Web开发中,前后端数据交互通常使用JSON格式。通过使用Json.NET的DeserializeObject方法,可以将接收到的JSON数据转换为C#对象,方便在后端进行处理和操作。 API调用:当使用第三方API进行数据交互时,常常会使用JSON格式进行数据传输。通过Js...
In JSON a literal string needs to be enclosed in quotation marks (either single or double). So, your 'test' is not valid JSON. A sampl of valid JSON would bevar test = "'12/14/2020'";(Note that after the equal sign there is a space, double quote, single quote, character '1',...
Json.NET is a popular high-performance JSON framework for .NET - Newtonsoft.Json/Src/Newtonsoft.Json.Tests/Serialization/TypeNameHandlingTests.cs at master · JamesNK/Newtonsoft.Json