Question

我的文档结构如下：

{'_id': 1, '_type': '2017-01-01',...}  --- (1)
{'_id': 1, '_type': '2017-01-02',...}  --- (2)
{'_id': 2, '_type': '2017-01-01',...}  --- (3)
{'_id': 2, '_type': '2017-01-02',...}  --- (4)

每个id可以属于不同的类型（这里是日期）。目的是获得与给定id列表匹配的文档具有最小值_type。

因此，对于id = [1,2]的输入，文档1＆amp;应返回3，因为它们具有_type（2017-01-01）的最小值。

这可以很容易地在每个id的基础上完成（通过查询给定id的最小可用类型），但这太贵了，因为给定的id列表在数千个范围内。

Answer 1

The following query would do what i need:

{
 "query": {
    "bool": {
      "must": [
        {
          "terms": {
            "OpID": [id_list]
          }
        }
      ]
    }
  },
  "aggs": {
    "distinct_id": {
      "terms": {
        "field": "OpID",
        "size": 100
      },
      "aggs": {
        "least_type": {
          "top_hits": {
            "sort": [{
              "as_of": {
                "order": "asc"
              }
            }
              ],
              "_source": {
                "includes": ["as_of", "OpID"]
              },
              "size": 1
          }
        }
      }
    }
  },
  "size": 0
}

Explanation: The bool filter filters out the documents with the given ID list. The aggregation part gives top 1 document found after arranging documents in order of the given field (as_of in this case). The meaning of _source remains the same here.

The above solution worked for me as i had a field as_of which is of date type and had the copy of the value stored in _type. The above aggregation isn't possible on the field _type

获取具有给定字段的最小值的文档

1 个答案: