Use this this site for representation: https://quicktype.io/csharp/ something like this may help you public class Item322A { public string prop1 { get; set; } public string prop2 { get; set; } public int prop3 { get; set; } public bool prop4 { get; set; } } public class Item2B { public string prop1 { get; … Read more
Groovy provides the inspect() method returns an object as a parseable string: // serialize def m = [a: 123, b: ‘test’] def str = m.inspect() // deserialize m = Eval.me(str) Another way to serialize a groovy map as a readable string is with JSON: // serialize import groovy.json.JsonBuilder def m = [a: 123, b: ‘test’] … Read more
Parameterless constructors need for any kind of deserialization. Imagine that you are implementing a deserializer. You need to: Get a type of object from the input stream (in this case it’s string) Instantiate the object. You have no way to do that if there is no default constructor. Read the properties/value from stream Assign the … Read more
Try this code: public class Receiver { public string receiver_tax_id { get; set;} public string total { get; set;} public string receiver_company_name { get; set;} public int status { get; set;} } And deserialize looks like follows: var result = JsonConvert.DeserializeObject<List<Receiver>>(responseString); var status = result[0].status;
Roll your own Conttract resolver like public class SignalRContractResolver : IContractResolver { private readonly Assembly assembly; private readonly IContractResolver camelCaseContractResolver; private readonly IContractResolver defaultContractSerializer; public SignalRContractResolver() { defaultContractSerializer = new DefaultContractResolver(); camelCaseContractResolver = new CamelCasePropertyNamesContractResolver(); assembly = typeof(Connection).Assembly; } public JsonContract ResolveContract(Type type) { if (type.Assembly.Equals(assembly)) { return defaultContractSerializer.ResolveContract(type); } return camelCaseContractResolver.ResolveContract(type); } } Register … Read more
The whole point of the Json() helper method is to serialize as JSON. If you want to return raw content, do that directly: return Content(jsonString, “application/json”);
Solution was to add .configure(DeserializationFeature.READ_DATE_TIMESTAMPS_AS_NANOSECONDS, false) to the ObjectMapper. Complete ObjectMapper looks like: ObjectMapper om = new ObjectMapper() .registerModule(new JavaTimeModule()) .configure(SerializationFeature.WRITE_DATE_TIMESTAMPS_AS_NANOSECONDS, false) .configure(SerializationFeature.FAIL_ON_EMPTY_BEANS, false) .configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false) .configure(DeserializationFeature.READ_DATE_TIMESTAMPS_AS_NANOSECONDS, false) .setSerializationInclusion(Include.NON_NULL);
Check the Length property of the stream. Length represents the number of bytes currently in the file. If it is 0, the file is empty.
It’s not really a deserialization error, it’s an ActiveRecord record-not-found error on a simple Model.find(id) query. If you want to know the details, log them in the delayed_job-2.1.3/lib/delayed/serialization/active_record.rb file, in the rescue statement, just before delayed-job stupidly raises the DeserializationError and throws the useful information away.
Make your tabID nullable: public int? tabID; Value null could not be deserialized to integer. That’s your 6-th part of json (or 5-th starting with 0): {\”bmk\”:\”test-m\”,\”state\”:\”on\”,\”type\”:\”motor\”,\”tabID\”:null} You could see, that tabID is null there.