我有一个产品,它有一个属性categoryIds。
"id" : 1,
"title" : "product",
"price" : "1100.00",
"categories" : [ the ids of the product's categories],
"tags" : [ the ids of the product's tags ],
"variants" : [ nested type with properties: name, definition, maybe in the future availability dates]我想根据查询中的类别对产品id进行分组。在POST _search中,我询问属于特定类别的产品(例如1,2,3),我也可以用变体限制它们。如何对我的答案进行分组/聚合,以获得类别的productIds列表?我想要得到的是:
{
"productsForCategories": {
"1": [
"product-1",
"product-2",
"product-3"
],
"2": [
"product-1",
"product-3",
"product-4"
],
"3": [
"product-5",
"product-6"
]
}
}提前感谢你的所有回答。
java生成了什么。
curl --location --request POST 'https://localhost:9200/products/_search' \
--header 'Content-Type: application/json' \
--data-raw '{
"size": 0,
"query": {
"bool": {
"must": [
{
"bool": {
"should": [
{
"term": {
"categories": {
"value": 7,
"boost": 1.0
}
}
}
],
"adjust_pure_negative": true,
"minimum_should_match": "1",
"boost": 1.0,
"_name": "fromRawQuery"
}
}
],
"filter": [
{
"bool": {
"adjust_pure_negative": true,
"boost": 1.0,
"_name": "filterPart"
}
}
],
"adjust_pure_negative": true,
"boost": 1.0,
"_name": "queryPart"
}
},
"_source": {
"includes": [
"categories",
"productType",
"relations"
],
"excludes": []
},
"stored_fields": "_id",
"sort": [
{
"_score": {
"order": "desc"
}
}
],
"aggregations": {
"agg": {
"global": {},
"aggregations": {
"categories": {
"terms": {
"field": "categories",
"size": 2147483647,
"min_doc_count": 1,
"shard_min_doc_count": 0,
"show_term_doc_count_error": false,
"order": [
{
"_count": "desc"
},
{
"_key": "asc"
}
]
},
"aggregations": {
"productsForCategories": {
"terms": {
"field": "_id",
"size": 2147483647,
"min_doc_count": 1,
"shard_min_doc_count": 0,
"show_term_doc_count_error": false,
"order": [
{
"_count": "desc"
},
{
"_key": "asc"
}
]
}
}
}
}
}
}
}
}'```发布于 2020-10-04 11:23:31
您可以使用terms aggregation,它是一个基于多存储桶值源的聚合,其中存储桶是动态构建的-每个唯一值一个。
添加一个包含索引数据、映射、搜索查询和搜索结果的工作示例
索引映射:
{
"mappings":{
"properties":{
"categories":{
"type":"keyword"
}
}
}
}索引数据:
{
"id":1,
"product":"p1",
"category":[1,2,7]
}
{
"id":2,
"product":"p2",
"category":[7,4,5]
}
{
"id":3,
"product":"p3",
"category":[4,5,6]
} 搜索查询:
{
"size": 0,
"aggs": {
"cats": {
"terms": {
"field": "cat_ids",
"include": [
7
]
},
"aggs": {
"products": {
"terms": {
"field": "product.keyword",
"size": 10
}
}
}
}
}
}搜索结果:
"aggregations": {
"cats": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": 7,
"doc_count": 2,
"products": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "p1",
"doc_count": 1
},
{
"key": "p2",
"doc_count": 1
}
]
}
}
]
}发布于 2020-10-04 14:31:14
我相信你想要的是与每个类别相对应的产品。正如Bhavya提到的,您可以使用term聚合来实现同样的功能。
GET products/_search
{
"size": 0, //<===== If you need only aggregated results, set this to 0. It represents query result size.
"aggs": {
"categories": {
"terms": {
"field": "cat_ids", // <================= Equivalent of group by Cat_ids
"size": 10
},"aggs": {
"products": {
"terms": {
"field": "name.keyword",//<============= For Each category group by products
"size": 10
}
}
}
}
}
}结果:
"aggregations" : {
"categories" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : 1, //<========== category id
"doc_count" : 2, //<========== For the given category id 2 products
"products" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : "p1", //<========= for cat_id=1, p1 is there
"doc_count" : 1
},
{
"key" : "p2", //<========= for cat_id=1, p2 is there
"doc_count" : 1
}
]
}
},
{
"key" : 2,
"doc_count" : 2,
"products" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : "p1",
"doc_count" : 1
},
{
"key" : "p2",
"doc_count" : 1
}
]
}
},
{
"key" : 3,
"doc_count" : 1,
"products" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : "p1",
"doc_count" : 1
}
]
}
}
]
}}
详细信息以注释的形式提供。请删除注释,然后尝试运行查询。
过滤聚合结果: See this
https://stackoverflow.com/questions/64189083
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