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[英]ElasticSearch Aggregation on nested field with bucketing on parent id
[英]How to sort nested aggregation field based on parent document field in elasticsearch?
我有不同位置的商店索引。 对于每家商店,我都有一个嵌套的折扣券列表。
现在我有查询以获取半径为 ax km 的所有唯一优惠券的列表,按给定位置上最近适用优惠券的距离排序
数据库:: Elasticsearch
索引映射::
{
"mappings": {
"car_stores": {
"properties": {
"location": {
"type": "geo_point"
},
"discount_coupons": {
"type": "nested",
"properties": {
"name": {
"type": "keyword"
}
}
}
}
}
}
}
样本文件::
{
"_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"
}
]
}
}
旧查询以获取给定位置 x 公里区域内的唯一折扣券名称::
{
"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
}
}
}
}
}
}
更新回复::
{
"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
}
}
}
]
}
}
}
}
现在上面的查询将返回按计数排序的前 200 张折扣优惠券默认值,但我想返回根据给定位置的距离排序的优惠券,即最近适用的优惠券应该排在第一位。
有没有办法根据父键对嵌套聚合进行排序,或者我可以使用不同的数据 model 来解决这个用例吗?
更新查询::
{
"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)"
}
}
}
}
}
}
}
}
}
}
}
有趣的问题。 您始终可以根据数字子聚合的结果对terms
聚合进行排序。 这里的技巧是通过reverse_nested
聚合来逃避嵌套上下文,然后使用脚本计算与 pivot 的距离:
{
"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)"
}
}
}
}
}
}
}
}
}
}
}
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