我正在使用
Java客户端API来获取聚合.以下是我正在处理的结构.
aggregations
top_models
buckets
key : "BMW"
doc_count : 3
top_models
buckets
key : "X5"
doc_count : 2
top_hits
source
model : X5
color : Red
source
model:X5
color : White
key : "X3"
doc_count : 1
top_hits
source
model : X3
color : Red
key : "Mercedes"
doc_count : 2
top_models
buckets
key : "Benz"
doc_count : 1
top_hits
source
model : Benz
color : Red
key : "ML"
doc_count : 1
top_hits
source
model : ML
color : Black
我正在尝试跟随(玩具)代码来检索所有结果.
def getAggregations(aggres: Option[Aggregations]): Option[Iterable[Any]] = {
aggres map { agg =>
val aggS = agg.asMap().asScala
aggS map {
case (name, termAgg: Terms) => getBuckets(Option(termAgg.getBuckets()))
case (name, topHits: TopHits) =>
val tHits = Option(topHits.getHits())
tHits map { th => getTopHits(th.asScala)
}
case (h, a: InternalAvg) => println(h + "=>" + a.getValue());
}
}
}
def getBuckets(buckets: Option[java.util.Collection[Bucket]]) = {
buckets map { bks =>
val bksS = bks.asScala
bksS map { b =>
println("Bucket Key =>" + b.getKey())
println("Doc count =>" + b.getDocCount())
getAggregations(Option(b.getAggregations())
}
}
}
需要将最终结果填充到此类
case class FinalResponse(bucketName: String, count: Long, children: List[FinalResponse])
通过聚合和桶之间的嵌套关系,检索所有聚合结果变得很复杂.你怎么看待这个?
最佳答案 在我之前的项目中,我们使用这种方式来预测弹性搜索中的复杂对象:
弹性搜索允许您获取结果集的每个元素的普通json视图(作为字符串).我们只使用带有scala模块的Jacson json库,并使类似pojo的类反序列化数据.
弹性搜索java api是一组可怕的多层嵌套映射.把它给忘了.
val hits = qb.execute().actionGet().getHits().getHits().asScala
hits.map { hit =>
(hit.getId, hit.getSourceAsString, hit.getVersion)
}
对于agregations,它也应该是可用的