I have the following class structure:
public class JsonModel
{
public string PropertyName { get; set; }
public string PropertyValue { get; set; }
}
And I have an instance of this class as follows:
var entry = new JsonModel { PropertyName = "foo", PropertyValue = "bar" };
I want to convert this to json, but I want it to come out like this:
{
"foo": "bar"
}
How cant his be done?
If you use Newtonsoft Json.NET you can use a JsonConverter implementation to handle it.
Here's a rather naive implementation:
public class JsonModelConverter : JsonConverter
{
public override bool CanConvert(Type objectType) => objectType == typeof(JsonModel);
public override object ReadJson(JsonReader reader, Type objectType,
object existingValue, JsonSerializer serializer)
{
var dict = serializer.Deserialize<Dictionary<string, string>>(reader);
var item = dict.First();
return new JsonModel { PropertyName = item.Key, PropertyValue = item.Value };
}
public override void WriteJson(JsonWriter writer, object value,
JsonSerializer serializer)
{
var model = (JsonModel)value;
var dict = new Dictionary<string, string>();
dict[model.PropertyName] = model.PropertyValue;
serializer.Serialize(writer, dict);
}
}
You have to pass this in to the SerializeObject and DeserializeObject<T> methods, here's a usage example with example output:
void Main()
{
var t = new Test
{
Model = new JsonModel { PropertyName = "key", PropertyValue = "some value" }
};
string json = JsonConvert.SerializeObject(t, new JsonModelConverter());
Console.WriteLine(json);
var t2 = JsonConvert.DeserializeObject<Test>(json, new JsonModelConverter());
Console.WriteLine($"{t2.Model.PropertyName}: {t2.Model.PropertyValue}");
}
public class Test
{
public JsonModel Model { get; set; }
}
Output:
{"Model":{"key":"some value"}}
key: some value
assuming that your class has only 2 properties ( name and value) you can use this
var entry = new JsonModel { PropertyName = "foo", PropertyValue = "bar" };
var json= GetJson(entry);
public string GetJson(object obj)
{
var prop= obj.GetType().GetProperties();
var prop1=prop[0].GetValue(obj, null).ToString();
var prop2= prop[1].GetValue(obj, null).ToString();
return "{\n\"" + $"{prop1}\": \"{prop2}" + "\"\n}";
}
json
{
"foo": "bar"
}
if you wnat to use a serializer, it can be done like this
public string GetJsonFromParsing(object obj)
{
var json=JsonConvert.SerializeObject(obj);
var properties=JObject.Parse(json).Properties().ToArray();
var prop1 = properties[0].Value;
var prop2 = properties[1].Value;
return "{\n\"" + $"{prop1}\": \"{prop2}" + "\"\n}";
}
var entry = new JsonModel { PropertyName = "fo\"o", PropertyValue = "bar" }; will result in { "fo"o": "bar" }, which is not a valid JSON.To output key value pairs, just specify what the key and value is in an interface. In your case:
public interface IKVP
{
string PropertyName { get; set; }
string PropertyValue { get; set; }
}
Then you can output any class that adheres to the interface:
public class JsonModel : IKVP
{
public string PropertyName { get; set; }
public string PropertyValue { get; set; }
}
Once that is done, create an extension method which will only take classes that implement the interface:
public static class ClassConverter
{
public static string ToJson<T>(this T value) where T : IKVP
=> $"{{ \x22{value.PropertyName}\x22 : \x22{value.PropertyValue}\x22 }}";
}
Once done, output the json string:
var entry = new JsonModel { PropertyName = "foo", PropertyValue = "bar" };
var json = entry.ToJson(); // { "foo" : "bar" }
Build upon this as needed for more complex situations.
var entry = new JsonModel { PropertyName = "fo\"o", PropertyValue = "bar" }; will result in { "fo"o": "bar" }, which is not a valid JSON.
ToJSON above to put in code to validate the json. That is why I put "Build upon this as needed for more complex situations." in my answer.in most programming languages, including C#, the common way to serialize to JSON would be to take the field's name as the key name, and the field's value as the value.
what you're asking to achieve is to somehow couple between two different fields, which is not a common thing to do.
StringBuilderand you're off to the races. But I guess your question is really how to do this automatically? In this case, you'll need to be more specific. Which serialization library are you using?