弹性搜索 - 从json响应中排除索引和类型 [英] Elastic Search - exclude index and type from json response

问题描述

当我通过这样的索引执行查询时：

When I execute a query over an index like this:

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

响应包括索引，类型等。但是我已经知道索引和类型，因为我指定它。这个信息只是膨胀了json数据的大小。有没有办法从回应中排除这些？

the response includes index, type, etc. But I already know the index and type because I specified it. This information just bloats up the size of the json data. Is there a way to exclude these from the response?

这是我得到的：

{
"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：

What I want is something like this, so a response without type,score,index:

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

推荐答案

是的，从ES 1.6起，您可以使用 filter_path 参数只枚举响应中需要的内容：

Yes, as of ES 1.6, you can use response filtering and using the filter_path parameter in the query enumerate only what you need in the response:

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

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