elasticsearch-获得“ function_score”内的中间分数

2024-05-03 • 问答

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))"
        }
      }
    }
  }
}

[
  {
    "_index": "blogs","_type": "1","_id": "AW5fxnperVbDy5wjSDBC","_score": 0.58024323,"_source": {
      "name": "learn elastic search","popularity": 100
    }
  },{
    "_index": "blogs","_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并没有考虑到受欢迎程度）

我探索过的东西：script_fields（在获取时无法访问_score。

aheiheiliu 回答：elasticsearch-获得“ function_score”内的中间分数

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上拉。

您可以尝试通过这种方式进行查询：

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

elasticsearch elasticsearch-aggregation search

本文链接：https://www.f2er.com/3117006.html

elasticsearch-获得“ function_score”内的中间分数

aheiheiliu 回答：elasticsearch-获得“ function_score”内的中间分数

大家都在问