ElasticSearch.NET NEST search<T> url

Question

My correct index path is POST: /foo/_search but below code hits POST: /foo/bar/_search.

var node = new Uri("http://elasticsearch-server.com:9200");
var settings = new ConnectionSettings(node);
settings.DefaultIndex("foo");
var client = new ElasticClient(settings);
var response = client.Search<Bar>(s => s
.Query(q => q.Term(o => o.userName, "test"))
);

// POCO for response fields
public class Bar
{
    public int userId { get; set; }
    public string userName { get; set; }
    public DateTime createdTime { get; set; }
}

Above code response returns below message;

Valid NEST response built from a successful low level call on POST: /foo/bar/_search

How can I set search path correctly?

Trial 1

When I omitted settings.DefaultIndex("foo"); line, it throws ArgumentException as below, but when I set DefaultIndex(), Search<T> uses T name as a second path.

ArgumentException: Index name is null for the given type and no default index is set. Map an index name using ConnectionSettings.MapDefaultTypeIndices() or set a default index using ConnectionSettings.DefaultIndex().

Trial 2 Refer to the documentation,

var settings = new ConnectionSettings(node)
.MapDefaultTypeIndices(m => m.Add(typeof(Bar), "foo"));

Above code returns same result in response.

Valid NEST response built from a successful low level call on POST: /foo/bar/_search

I have the same code and it works well. could you try this snippet of the code: var settings = new ConnectionSettings(new Uri("elasticsearch-server.com:9200")).DefaultIndex("foo&qu… client = new ElasticClient(settings); — Søren
– Søren, Commented Jan 8, 2017 at 8:30
@Soren // Thanks for your comment. Hmm... still has same issue. It uses POCO name as a second path. I need to investigate more. — Youngjae
– Youngjae, Commented Jan 8, 2017 at 8:54

Russ Cam · Accepted Answer · 2017-01-09 00:01:02Z

A large proportion of the Elasticsearch API exposed through NEST is in a strongly typed fashion, including .Search<T>(); with this endpoint, both "index" and "type" will be inferred from T, but sometimes you might want to set a different value to that which is inferred. In these cases, you can call additional methods on the search fluent API (or search object, if using the object initializer syntax) to override the inferred values

void Main()
{
    var pool = new SingleNodeConnectionPool(new Uri("http://localhost:9200"));
    var connectionSettings = new ConnectionSettings(pool)
            .DefaultIndex("foo");

    var client = new ElasticClient(connectionSettings);

    // POST http://localhost:9200/foo/bar/_search
    // Will try to deserialize all _source to instances of Bar
    client.Search<Bar>(s => s
        .MatchAll()
    );

    // POST http://localhost:9200/foo/_search
    // Will try to deserialize all _source to instances of Bar
    client.Search<Bar>(s => s
        .AllTypes()
        .MatchAll()
    );

    // POST http://localhost:9200/_search
    // Will try to deserialize all _source to instances of Bar
    client.Search<Bar>(s => s
        .AllTypes()
        .AllIndices()
        .MatchAll()
    );

    connectionSettings = new ConnectionSettings(pool)
            .InferMappingFor<Bar>(m => m
                .IndexName("bars")
                .TypeName("barbar")
            );

    client = new ElasticClient(connectionSettings);

    // POST http://localhost:9200/bars/barbar/_search
    // Will try to deserialize all _source to instances of Bar
    client.Search<Bar>(s => s
        .MatchAll()
    );

    // POST http://localhost:9200/bars/_search
    // Will try to deserialize all _source to instances of Bar
    client.Search<Bar>(s => s
        .AllTypes()
        .MatchAll()
    );

    // POST http://localhost:9200/_all/barbar/_search
    // Will try to deserialize all _source to instances of Bar
    client.Search<Bar>(s => s
        .AllIndices()
        .MatchAll()
    );

    // POST http://localhost:9200/_search
    // Will try to deserialize all _source to instances of Bar
    client.Search<Bar>(s => s
        .AllIndices()
        .AllTypes()
        .MatchAll()
    );
}


public class Bar
{
    public int userId { get; set; }
    public string userName { get; set; }
    public DateTime createdTime { get; set; }
}

tottishi05 · Accepted Answer · 2017-01-08 09:34:24Z

7

You may add other parameters in your search lambda expression var response = client.Search<Bar>(s => s.Index("indexName").Query(q => q.Term(o => o.userName, "test")));

answered Jan 8, 2017 at 9:34

tottishi05

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2 Comments

Youngjae Over a year ago

Hmm...either Index("indexName") or DefaultIndex("indexName") gives identical path.

Russ Cam Over a year ago

.DefaultIndex("defaultIndex") will be used if no index name is specified in the request, and no index name is set to be inferred for a given type T. See Index Name Inference for more details: elastic.co/guide/en/elasticsearch/client/net-api/5.x/…

Youngjae · Accepted Answer · 2017-01-08 13:00:36Z

0

I was newbie in ElasticSearch and didn't know about _type.

I set identical _type name to POCO class name, and it works as I expected. So we can say, {index}/{type} is what path expression is.

answered Jan 8, 2017 at 13:00

Youngjae

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