弹性搜索 - 从json响应中排除索引和类型

Question

当我对这样的索引执行查询时:

{
   "_source":["bar"] , "size":100,
   "query": {
       "match_all": {}
   },
   "filter": {
       "type" : {
           "value" : "foo"
        }
    }
}

响应包括索引,类型等.但我已经知道索引和类型,因为我指定了它.这些信息只会膨胀json数据的大小.有没有办法从响应中排除这些？

这就是我得到的:

{
"took": 31,
"timed_out": false,
"_shards": {
    "total": 5,
    "successful": 5,
    "failed": 0
    },
"hits": {
    "total": 364024,
    "max_score": 1,
    "hits": [
          {
        "_index": "foo_bar",
        "_type": "foo",
        "_id": "asdjj123123",
        "_score": 1,
        "_source": {
          "bar": "blablablabla"
    }
  }
,...

我想要的是这样的,所以没有类型,分数,索引的响应:

{
"took": 31,
"timed_out": false,
"_shards": {
    "total": 5,
    "successful": 5,
    "failed": 0
    },
"hits": {
    "total": 364024,
    "max_score": 1,
    "hits": [
          {
        "_id": "asdjj123123",
        "_source": {
          "bar": "blablablabla"
    }
  }
,...

Answer 1

是的,从ES 1.6开始,您可以使用响应过滤并使用filter_path查询中的参数仅枚举响应中所需的内容:

curl -XGET 'localhost:9200/foo_bar/foo/_search?pretty&filter_path=hits.total,hits.max_score,hits.hits._id,hits.hits._source'