我想输出弹性搜索的每个结果的分数。但我不确定如何才能得到这个。
以下是我当前运行查询的代码:
var searchResults = client.Search<Place>(s => s
.From(0)
.Size(5)
.Explain(true)
.TrackScores(true)
.Query(q => q
.QueryString(fqqs1 => fqqs1
.OnFieldsWithBoost(d => d
.Add("name", 5.0)
)
.Query("west midlands birmingham")
)
)
.Sort(sort => sort.OnField("_score").Descending())
.Sort(sort => sort.OnField(f => f.id).Ascending())
);
// Output the results to console
Console.WriteLine("\nTotal Hits: " + searchResults.HitsMetaData.Hits.Count + " out of " + searchResults.HitsMetaData.Total);
List<Result> results = new List<Result>();
foreach (Place result in searchResults.Documents)
{
results.Add(new Result
{
woeid = Convert.ToInt32(result.id),
name = result.name,
admin1 = result.admin1,
admin2 = result.admin2,
type = result.type
});
Console.WriteLine(result.id + " > " + result.name + " > " + result.admin1 + " > " + result.admin2 + " > " + result.type);
}
答案 0 :(得分:3)
使用.Hits
上的ISearchResponse<T>
属性集合 - 该集合包含.Score
属性中每个文档的分数,以及.Source
属性中的文档
答案 1 :(得分:0)
您可以按分数排序。例如:排序(sort => sort.OnField(“ _ score”)。Descending())
var result = client.Search(q => q
.Index(your-index-name)
.From(0)
.Type("post")
.Fields("firstName","LastName")
.TrackScores(true)
.Size(12)
.Query(SearchQuery)
.Sort(sort => sort.OnField("_score").Descending())
);
Nest 7的代码示例
var sorts = new List<ISort>();
sorts.Add(new FieldSort { Field = "_score", Order = SortOrder.Descending });
var searchRequest = new SearchRequest<ElasticIndexGroupProduct>()
{
Profile = true,
From = (pageNumber - 1) * pageSize,
Size = pageSize,
Version = true,
Sort = sorts,
Query = new MatchAllQuery()
Aggregations = aggrigations
};
var searchResponse = _client.Search<ElasticIndexGroupProduct>(searchRequest);