C# MongoDB kompleks klasse serialisering

Fandt hvordan du gemmer dataene til MongoDB her:Ordbog-til-BsonDocument-konvertering uden _t-felt og udvidede det lidt, så jeg tænkte at dele den fulde løsning.

Trin #1:

I min klasse erklærede jeg 2 medlemmer for hver værdi:

// For the Hobbies object type:
[BsonIgnore] //ignore this value in MongoDB
public Dictionary<string, object> Hobbies { get; set; }

[JsonIgnore] //ignore this value in the response on Get requests
[BsonElement(elementName: "Hobbies")]
public BsonDocument HobbiesBson { get; set; }

/*********************************************************************/

// For the Collection object type:
[BsonIgnore] //ignore this value in MongoDB
public List<Dictionary<string, object>> Collection { get; set; }

[JsonIgnore] //ignore this value in the response on Get requests
[BsonElement(elementName: "Collection")]
public BsonArray CollectionBson { get; set; }

Trin #2

I min WebAPI-controller metode til Post

[HttpPost]
public override async Task<IActionResult> Post([FromBody] Person person)
{
    var jsonDoc = JsonConvert.SerializeObject(person.Hobbies);
    person.HobbiesBson = BsonSerializer.Deserialize<BsonDocument>(jsonDoc);

    jsonDoc = JsonConvert.SerializeObject(person.Collection);
    person.CollectionBson = BsonSerializer.Deserialize<BsonArray>(jsonDoc);

    //save
}

Trin #3

I min Get anmod om, at jeg deserialiserer det tilbage sådan her:

[HttpGet("{id?}")]
public override async Task<IActionResult> Get(string id = null)
{
    var people = //get data from mongoDB
    foreach (var person in people)
    {
        var bsonDoc = BsonExtensionMethods.ToJson(person.HobbiesBson);
        person.Hobbies = JsonConvert.DeserializeObject<Dictionary<string, object>>(bsonDoc);

        bsonDoc = BsonExtensionMethods.ToJson(person.CollectionBson);
        person.Collection = JsonConvert.DeserializeObject<List<Dictionary<string, object>>>(bsonDoc);bsonDoc);
    }
    return Ok(people);
}

Dette løste mit problem, og jeg håber, det kan hjælpe andre også :-)