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elasticsearch - elasticsearch - 在“function_score”中获得中间分数

问题描述

这是我的索引

POST /blogs/1
{
  "name" : "learn java", 
  "popularity" : 100
}

POST /blogs/2
{
  "name" : "learn elasticsearch", 
  "popularity" : 10
}

我的搜索查询：

GET /blogs/_search
{
  "query": {
    "function_score": {
      "query": {
        "match": {
          "name": "learn"
        }
      },
      "script_score": {
        "script": {
          "source": "_score*(1+Math.log(1+doc['popularity'].value))"
        }
      }
    }
  }
}

返回：

[
  {
    "_index": "blogs",
    "_type": "1",
    "_id": "AW5fxnperVbDy5wjSDBC",
    "_score": 0.58024323,
    "_source": {
      "name": "learn elastic search",
      "popularity": 100
    }
  },
  {
    "_index": "blogs",
    "_type": "1",
    "_id": "AW5fxqmL8cCMCxtBYOyC",
    "_score": 0.43638366,
    "_source": {
      "name": "learn java",
      "popularity": 10
    }
  }
]

问题：我需要在结果中返回一个额外的字段，这会给我原始分数（只是 tf/idf 不考虑流行度）

我探索过的东西：script_fields_score （在获取时不提供访问权限。

标签： elasticsearchsearchelasticsearch-aggregation

解决方案

问题在于您查询的方式，它会覆盖_score变量。相反，如果您使用sortthen_score不会更改并且可以在同一查询中提取。

您可以尝试这样查询：

{
  "query": {
    "match": {
      "name": "learn"
    }
  },
  "sort": [
    {
      "_script": {
        "type": "number",
        "script": {
          "lang": "painless",
          "source": "_score*(1+Math.log(1+doc['popularity'].value))"
        },
        "order": "desc"
      }
    },
    "_score"
  ]
}

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