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Getting a single object from mongodb in C#

I've picked up a piece of code that is using the MongoDB driver like this to get a single object from a collection...this can't be right, can it? Is there a better way of getting this?

IMongoCollection<ApplicationUser> userCollection;
....
userCollection.FindAsync(x => x.Id == inputId).Result.ToListAsync().Result.Single();
over 4 years ago · Santiago Trujillo
3 Respostas
Responde à pergunta

0

Yes, there is.

First of all don't use FindAsync, use Find instead. On the IFindFluent result use the SingleAsync extension method and await the returned task inside an async method:

async Task MainAsync()
{
    IMongoCollection<ApplicationUser> userCollection = ...;

    var applicationUser = await userCollection.Find(_ => _.Id == inputId).SingleAsync();
}

The new driver uses async-await exclusively. Don't block on it by using Task.Result.

over 4 years ago · Santiago Trujillo Relatório

0

You should limit your query before executing, otherwise you will first find all results and then only read one of it.

You could either specify the limit using FindOptions in FindAsync, or use the fluent syntax to limit the query before executing it:

var results = await userCollection.Find(x => x.Id == inputId).Limit(1).ToListAsync();
ApplicationUser singleResult = results.FirstOrDefault();

The result from ToListAsync will be a list but since you limited the number of results to 1, that list will only have a single result which you can access using Linq.

over 4 years ago · Santiago Trujillo Relatório

0

In newer versions of MongoDB Find() is deprecated, so you can either use

collection.FindSync(o => o.Id == myId).Single()

or

collection.FindAsync(o => o.Id == myId).Result.Single()

You can also use SingleOrDefault(), which returns null if no match was found instead of throwing an exception.

over 4 years ago · Santiago Trujillo Relatório
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