Question

我的正确索引路径为POST: /foo/_search，但代码点击POST: /foo/bar/_search以下。

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; }
}

以上代码response会返回以下消息;

在POST上成功进行低级别调用构建的有效NEST响应：/ foo / bar / _search

如何正确设置搜索路径？

试用1

当我省略settings.DefaultIndex("foo");行时，它会抛出ArgumentException，如下所示，但当我设置DefaultIndex()时，Search<T>使用T名称作为第二条路径。

ArgumentException：给定类型的索引名称为null，并且未设置默认索引。使用ConnectionSettings.MapDefaultTypeIndices（）映射索引名称，或使用ConnectionSettings.DefaultIndex（）设置默认索引。

试用2 请参阅documentation，

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

上面的代码在响应中返回相同的结果。

在POST上成功进行低级别调用构建的有效NEST响应：/ foo / bar / _search

Answer 1

通过NEST公开的大部分Elasticsearch API采用强类型方式，包括.Search<T>();使用此端点，"index"和"type"都将从T推断，但有时您可能希望为推断的值设置不同的值。在这些情况下，您可以在搜索流畅的API（或搜索对象，如果使用对象初始化程序语法）上调用其他方法来覆盖推断的值

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; }
}

Answer 2

您可以在搜索lambda表达式中添加其他参数 var response = client.Search<Bar>(s => s.Index("indexName").Query(q => q.Term(o => o.userName, "test")));

Answer 3

我是ElasticSearch的新手，并不了解_type。

我将相同的_type名称设置为POCO类名称，它按预期工作。所以我们可以说，{index}/{type}是路径表达式。

ElasticSearch.NET NEST搜索<t> url

3 个答案: