如何根據 elasticsearch 中的父文檔字段對嵌套聚合字段進行排序？

Question

我有不同位置的商店索引。 對於每家商店，我都有一個嵌套的折扣券列表。

現在我有查詢以獲取半徑為 ax km 的所有唯一優惠券的列表，按給定位置上最近適用優惠券的距離排序

數據庫:: Elasticsearch

索引映射::

{
"mappings": {
    "car_stores": {
        "properties": {
            "location": {
                "type": "geo_point"
            },
            "discount_coupons": {
                "type": "nested",
                "properties": {
                    "name": {
                        "type": "keyword"
                    }
                }
            }
        }
    }
}
}

樣本文件::

{
"_index": "stores",
"_type": "car_stores",
"_id": "1258c81d-b6f2-400f-a448-bd728f524b55",
"_score": 1.0,
"_source": {
    "location": {
        "lat": 36.053757,
        "lon": 139.525482
    },
    "discount_coupons": [
        {
            "name": "c1"
        },
        {
            "name": "c2"
        }
    ]
}
}

舊查詢以獲取給定位置 x 公里區域內的唯一折扣券名稱::

{
"size": 0,
"query": {
    "bool": {
        "must": {
            "match_all": {}
        },
        "filter": {
            "geo_distance": {
                "distance": "100km",
                "location": {
                    "lat": 40,
                    "lon": -70
                }
            }
        }
    }
},
"aggs": {
    "coupon": {
        "nested": {
            "path": "discount_coupons"
        },
        "aggs": {
            "name": {
                "terms": {
                    "field": "discount_coupons.name",
                    "order": {
                        "_key": "asc"
                    },
                    "size": 200
                }
            }
        }
    }
}
}

更新回復::

{
"took": 60,
"timed_out": false,
"_shards": {
    "total": 3,
    "successful": 3,
    "skipped": 0,
    "failed": 0
},
"hits": {
    "total": 245328,
    "max_score": 0.0,
    "hits": []
},
"aggregations": {
    "coupon": {
        "doc_count": 657442,
        "name": {
            "doc_count_error_upper_bound": -1,
            "sum_other_doc_count": 641189,
            "buckets": [
                {
                    "key": "local20210211",
                    "doc_count": 1611,
                    "back_to_base": {
                        "doc_count": 1611,
                        "distance_script": {
                            "value": 160.61034409639765
                        }
                    }
                },
                {
                    "key": "local20210117",
                    "doc_count": 1621,
                    "back_to_base": {
                        "doc_count": 1621,
                        "distance_script": {
                            "value": 77.51459886447356
                        }
                    }
                },
                {
                    "key": "local20201220",
                    "doc_count": 1622,
                    "back_to_base": {
                        "doc_count": 1622,
                        "distance_script": {
                            "value": 84.15734462544432
                        }
                    }
                },
                {
                    "key": "kisekae1",
                    "doc_count": 1626,
                    "back_to_base": {
                        "doc_count": 1626,
                        "distance_script": {
                            "value": 88.23770888201268
                        }
                    }
                },
                {
                    "key": "local20210206",
                    "doc_count": 1626,
                    "back_to_base": {
                        "doc_count": 1626,
                        "distance_script": {
                            "value": 86.78376012847237
                        }
                    }
                },
                {
                    "key": "local20210106",
                    "doc_count": 1628,
                    "back_to_base": {
                        "doc_count": 1628,
                        "distance_script": {
                            "value": 384.12156408078397
                        }
                    }
                },
                {
                    "key": "local20210113",
                    "doc_count": 1628,
                    "back_to_base": {
                        "doc_count": 1628,
                        "distance_script": {
                            "value": 153.61681676703674
                        }
                    }
                },
                {
                    "key": "local20",
                    "doc_count": 1629,
                    "back_to_base": {
                        "doc_count": 1629,
                        "distance_script": {
                            "value": 168.74132991524073
                        }
                    }
                },
                {
                    "key": "local20210213",
                    "doc_count": 1630,
                    "back_to_base": {
                        "doc_count": 1630,
                        "distance_script": {
                            "value": 155.8335679860034
                        }
                    }
                },
                {
                    "key": "local20210208",
                    "doc_count": 1632,
                    "back_to_base": {
                        "doc_count": 1632,
                        "distance_script": {
                            "value": 99.58790590445102
                        }
                    }
                }
            ]
        }
    }
}
}

現在上面的查詢將返回按計數排序的前 200 張折扣優惠券默認值，但我想返回根據給定位置的距離排序的優惠券，即最近適用的優惠券應該排在第一位。

有沒有辦法根據父鍵對嵌套聚合進行排序，或者我可以使用不同的數據 model 來解決這個用例嗎？

更新查詢::

{
"size": 0,
"query": {
    "bool": {
        "filter": [
            {
                "geo_distance": {
                    "distance": "100km",
                    "location": {
                        "lat": 35.699104,
                        "lon": 139.825211
                    }
                }
            },
            {
                "nested": {
                    "path": "discount_coupons",
                    "query": {
                        "bool": {
                            "filter": {
                                "exists": {
                                    "field": "discount_coupons"
                                }
                            }
                        }
                    }
                }
            }
        ]
    }
},
"aggs": {
    "coupon": {
        "nested": {
            "path": "discount_coupons"
        },
        "aggs": {
            "name": {
                "terms": {
                    "field": "discount_coupons.name",
                    "order": {
                        "back_to_base": "asc"
                    },
                    "size": 10
                },
                "aggs": {
                    "back_to_base": {
                        "reverse_nested": {},
                        "aggs": {
                            "distance_script": {
                                "min": {
                                    "script": {
                                        "source": "doc['location'].arcDistance(35.699104, 139.825211)"
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }
}
}

Answer 1

有趣的問題。 您始終可以根據數字子聚合的結果對terms聚合進行排序。 這里的技巧是通過reverse_nested聚合來逃避嵌套上下文，然后使用腳本計算與 pivot 的距離：

{
  "size": 0,
  "query": {
    "bool": {
      "must": {
        "match_all": {}
      },
      "filter": {
        "geo_distance": {
          "distance": "100km",
          "location": {
            "lat": 40,
            "lon": -70
          }
        }
      }
    }
  },
  "aggs": {
    "coupon": {
      "nested": {
        "path": "discount_coupons"
      },
      "aggs": {
        "name": {
          "terms": {
            "field": "discount_coupons.name",
            "order": {
              "back_to_base": "asc"
            },
            "size": 200
          },
          "aggs": {
            "back_to_base": {
              "reverse_nested": {},
              "aggs": {
                "distance_script": {
                  "min": {
                    "script": {
                      "source": "doc['location'].arcDistance(40, -70)"
                    }
                  }
                }
              }
            }
          }
        }
      }
    }
  }
}