Scala 如何从ElasticSearch检索存储的可变集
嗨,我用这种方式在弹性搜索中选择一个id字段和一个Scala可变集Scala 如何从ElasticSearch检索存储的可变集,scala,
elasticsearch,scala-2.11,Scala,
elasticsearch,Scala 2.11,嗨,我用这种方式在弹性搜索中选择一个id字段和一个Scala可变集 var genreIdSet = scala.collection.mutable.Set[Int]() genreIdSet+=1 genreIdSet+=2 genreIdSet+=3 bulkRequest.add(client.prepareIndex("testdb","test","123") .setSource(jsonBuilder() .startObj
var genreIdSet = scala.collection.mutable.Set[Int]()
genreIdSet+=1
genreIdSet+=2
genreIdSet+=3
bulkRequest.add(client.prepareIndex("testdb","test","123")
.setSource(jsonBuilder()
.startObject()
.field("uuid","123")
.field("GenreIdSet",genreIdSet)
.endObject()
)
)
现在我想检索这个文档,这里是代码
val get=client.prepareGet("testdb","test","123")
.setOperationThreaded(false)
.setFields("uuid","GenreIdSet")
.execute()
.actionGet()
id=get.getField("uuid").getValue.toString().toInt
var a=get.getField("GenreIdSet").getValue.toString
id=get.getField("uuid").getValue.toString().toInt
var a=get.getField("GetGenreIdSet").getValues.toArray()
for(number<-a)
{
genreIdSet+=number.toString().toInt
}
我得到以下输出
ID is 123
GenreIdSet is Set(1, 2, 3)
我想遍历此集合并将其值存储在一个新的scala可变集合中,例如:1 2 3如何实现此目的请帮助我也是elasticSearch的新手,我正在接受elasticSearch java api的帮助。请帮助。谢谢。您可以使用数组方法在elasticSearch中插入一个scala可变集合
var genreIdSet = scala.collection.mutable.Set[Int]()
genreIdSet+=1
genreIdSet+=2
genreIdSet+=3
var xb:XContentBuilder=XContentFactory.jsonBuilder().startObject().field("uuid",artistImpl.getUuid)
xb.startArray("GetGenreIdSet")
for(n<-genreIdSet)
{
xb.value(n)
}
xb.endArray()
xb.endObject()
val bulkRequest=client.prepareBulk()
bulkRequest.add(client.prepareIndex("testdb","test","123")
.setSource(xb)
)
检索此数组并将其值存储到新的scala可变集下面是代码
val get=client.prepareGet("testdb","test","123")
.setOperationThreaded(false)
.setFields("uuid","GenreIdSet")
.execute()
.actionGet()
id=get.getField("uuid").getValue.toString().toInt
var a=get.getField("GenreIdSet").getValue.toString
id=get.getField("uuid").getValue.toString().toInt
var a=get.getField("GetGenreIdSet").getValues.toArray()
for(number<-a)
{
genreIdSet+=number.toString().toInt
}