我正在进行下面的通配符搜索,我想让(她的*)结果的开头比包含(她的)搜索结果的排名更多。有没有办法实现这个目标?
elasticClient.Search<ClientSearchContract>(p =>
p.Query(q => q
.Bool(bq => bq
.Should(mq => mq.QueryString(qs =>
qs.Query(her* *her*))
)
)
)
.Sort(x => x.OnField("_score").Descending())
);
答案 0 :(得分:0)
为此,我们使用custom score脚本。在JSON中,它看起来像这样:
"query":
{
"custom_score":
{
"query": *query* ,
"params" :
{
"streetMultiply": " + ((!searchQuery.HasCityValue()) ? 100 : 0) + @",
"addressMultiply": " + ((!searchQuery.HasStreetValue()) ? 100 : 0) + @"
}
,"script": "if (doc['streetId'].value == 0) { _score * streetMultiply } else { if (doc['addressId'].value == 0) { _score * addressMultiply } else { _score } }"
}
}, "size": 100 }"
我们没有使用NEST,但我在NEST找到了pull-requst来实现这个
答案 1 :(得分:0)
这可以帮到你:
var result = elasticClient.Search<ClientSearchContract>(q => q
.Index("index")
.Type("type")
.Query(qq =>
{
QueryContainer startsQuery = null;
QueryContainer containQuery = null;
{
startsQuery |= qq.QueryString(qs => qs.OnFields(f => f.Field).Query("her*").Boost(2));
containQuery |= qq.QueryString(qs => qs.OnFields(f => f.Field).Query("*her*").Boost(1.2));
}
return startsQuery || containQuery;
})
.Take(10)
.Sort(x => x.OnField("_score").Descending())
);