elasticsearch - elasticsearch-获得 'function_score'内的中级分数

Question

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，这将给我原始分数(只是tf / idf并没有考虑受欢迎程度)

我研究过的内容: script_fields(在获取时不提供对 _score的访问。

Answer 1

问题出在您查询的方式上，它覆盖了_score变量。相反，如果您使用sort，则_score不会更改，并且可以在同一查询中提取。

您可以尝试以这种方式查询:

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