You are missing quotes around the key names… “name”: “Samsung Tab 3”, The main issue here is the formatting in the description. You will get issues with syntax like: “Samsung Tab 3 is very well ” + “Samsung Tab 3 is very well “, Take a look at JSONLint too, this will help validate your … Read more
Step 1. Make some models (or Plain Old C# Classes if you like) to deserialize your json into (generated by http://json2csharp.com/) – this is generally a little bit easier to work with that straight up json: public class Customise { public string name { get; set; } public int id { get; set; } public … Read more
string json = “{Table1: [{\”PropertyTaxReceiptAmt\”:\”2200170.00\”,\”WaterTaxReceiptAmt\”:\”79265.00\”}]}”; json = json.Replace(“\\”, string.Empty); json = json.Trim(‘”‘); json = json.Replace(“{Table1:”, string.Empty); json = json.Remove(json.Length -1,1); Console.Write(json); 1 solved Remove character from json string
There is no question here but I am assuming you want to serialize that json string into an object? Also I would at least post what you have tried and what errors you are receiving if any. This is per the JSON.net web site found here http://james.newtonking.com/pages/json-net.aspx string json = “<you json string>”; JObject o … Read more
class Employer { [JsonProperty(“nome”)] public string Nome { get; set; } [JsonProperty(“id”)] public string Id { get; set; } [JsonProperty(“setor”)] public string Setor { get; set; } } class Employers : List<Employer> { } Employers employers = JObject.Parse(json)[“result”][“employers”].ToObject<Employers>(); 1 solved How to deserialize JSON in C# [closed]
Nodes is a JArray, not a JObject, so cannot be casted as such. Try this: var json = JsonConvert.DeserializeObject<dynamic>(JsonData); var nodes = ((JArray)json.Nodes).ToObject<string[]>(); solved C# JSON Variable Name reserved [closed]
Managed to get it working using JObject but it’s not pretty: var json = “{\r\n\”Threads\”: \r\n{\r\n \”Program1\” : \r\n {\r\n \”Filepath\”: \”C:\\\\ProgramFiles(x86)…\”,\r\n \”Priority\”: 0\r\n },\r\n \”Program2\” : \r\n {\r\n \”Filepath\”: \”C:\\\\ProgramFiles(x86)…\”,\r\n \”Priority\”: 1\r\n },\r\n \”Program3\” : \r\n {\r\n \”Filepath\”: \”C:\\\\ProgramFiles(x86)…\”,\r\n \”Priority\”: 3\r\n }\r\n}}”; var obj = JObject.Parse(json); var threads = (JObject)obj[“Threads”]; var sortedObj = new … Read more
You need to use the generic overload JsonSerializer.Deserialize<T>() var root = serializer.Deserialize<API_Json_Special_Feeds.RootObject>(jsonTextReader); Unlike files generated by BinaryFormatter, JSON files generally do not include c# type information, so it’s necessary for the receiving system to specify the expected type. (There are extensions to the JSON standard in which c# type information can be included in a … Read more
I solved it like this way: var json = JObject.Parse(responseString); Console.WriteLine(json[“data”][“authentication_token”]); solved Need to parse this data with Json.NET
Serializing a List<LinkedListNode<string>> is a bit of an odd thing to do – normally one would just serialize the underlying linked list. Perhaps you’re trying to serialize a table that gives the nodes in a different order than the underlying list? If that’s the case, it might appear that one could serialize the node list … Read more
In JSON format, numbers and booleans do not have quotes around them, while strings do (see JSON.org). If you want quotes around all your primitives, you have a few options: Change the properties of the object you are serializing to be of type string. Put your values into a Dictionary<string, string> (or a DTO) and … Read more
You can use a dictionary to parse a JSON object with custom property names: public class Test { public string col1 { get; set; } public string col2 { get; set; } } public class DataValue { public Test test { get; set; } } public class RootObject { public RootObject() { this.data = new … Read more
Create C# classes that map to your json. As below: public class Data { public List<Employee> Employees { get; set; } } public class Employee { public List<EmpDetails> EmpDetailss { get; set; } } public class EmpDetails { public int Empid { get; set; } public string Empname { get; set; } public string Empdept … Read more