我在每个产品的弹性搜索中都有一个单独的文档。 每个产品都有唯一 productId 和非唯一 groupId 以及其他属性，例如： categories < / em>的
我希望能够将不同的属性与每个唯一 groupId的计数进行汇总
示例：
doc 1：
{
"productId": 123
"groupId" xyz,
"categories": [{"value": "shoes"}, {"value": "t-shirt"}]
}
doc 2：
{
"productId": 345
"groupId" xyz,
"categories": [{"value": "shoes"}, {"value": "t-shirt"}]
}
doc 3：
{
"productId": 456
"groupId" abc,
"categories": [{"value": "t-shirt"}]
}
doc 4：
{
"productId": 567
"groupId" abc,
"categories": [{"value": "shoes"}, {"value": "makeup"}]
}
预期结果，如：
所以如果每个项目存在同一个 groupId
，我希望计数一次我的疑问：
{
"from":0,
"size":0,
"query":{
"filtered":{
"filter":{
}
}
},
"aggs": {
"group": {
"terms": {"field": "group"},
"aggs": {
"brand": {
"terms": {"field": "productMeta.brand.value"}
}
}
}
}
}
回复：
{
"took": 6,
"timed_out": false,
"_shards": {
"total": 5,
"successful": 5,
"failed": 0
},
"hits": {
"total": 25,
"max_score": 0,
"hits": []
},
"aggregations": {
"group": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 3,
"buckets": [
{
"key": "wlmr34210507",
"doc_count": 8,
"brand": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "generic",
"doc_count": 8
}
]
}
},
{
"key": "wlmr19524441",
"doc_count": 4,
"brand": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "maybelline",
"doc_count": 4
}
]
}
},
{
"key": "wlmr34121549",
"doc_count": 2,
"brand": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "maybelline",
"doc_count": 2
}
]
}
},
{
"key": "wlmr34317301",
"doc_count": 2,
"brand": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "dream on me",
"doc_count": 2
}
]
}
},
{
"key": "bbfs40549552",
"doc_count": 1,
"brand": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "samsung",
"doc_count": 1
}
]
}
},
{
"key": "bobb7937347",
"doc_count": 1,
"brand": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "chicco",
"doc_count": 1
}
]
}
},
{
"key": "wlmr24241413",
"doc_count": 1,
"brand": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "maybelline",
"doc_count": 1
}
]
}
},
{
"key": "wlmr27504560",
"doc_count": 1,
"brand": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "mr. beer",
"doc_count": 1
}
]
}
},
{
"key": "wlmr33986448",
"doc_count": 1,
"brand": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "mr. beer",
"doc_count": 1
}
]
}
},
{
"key": "wlmr40806575",
"doc_count": 1,
"brand": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "healthtex",
"doc_count": 1
}
]
}
}
]
}
}
}
答案 0 :(得分：1)
所以基本上我能够通过使用基数来解决这个问题，如下所示：
{
"from":0,
"size":0,
"query":{
"filtered":{
"filter":{
}
}
},
"sort":{
"ts":{
"order":"desc",
"mode":"max",
"ignore_unmapped":true
}
},
"aggs":{
"categories":{
"terms":{
"field":"productMeta.brand.value",
"size":0
},
"aggs": {
"category" : {
"cardinality" : {
"field" : "group"
}
}
}
}
}
}
每个 productId
{
"took": 4,
"timed_out": false,
"_shards": {
"total": 5,
"successful": 5,
"failed": 0
},
"hits": {
"total": 71,
"max_score": 0,
"hits": []
},
"aggregations": {
"categories": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "chocolate",
"doc_count": 23,
"category": {
"value": 23
}
},
{
"key": "notebook",
"doc_count": 9,
"category": {
"value": 1
}
},
{
"key": "olive_oil",
"doc_count": 7,
"category": {
"value": 7
}
},
{
"key": "physical_training",
"doc_count": 5,
"category": {
"value": 5
}
},
{
"key": "ski",
"doc_count": 5,
"category": {
"value": 2
}
},
{
"key": "gym_membership",
"doc_count": 4,
"category": {
"value": 4
}
},
{
"key": "ski_boots",
"doc_count": 4,
"category": {
"value": 1
}
},
{
"key": "vinegar",
"doc_count": 4,
"category": {
"value": 4
}
},
{
"key": "bracelet",
"doc_count": 3,
"category": {
"value": 3
}
},
{
"key": "handbags",
"doc_count": 2,
"category": {
"value": 2
}
},
{
"key": "cider",
"doc_count": 1,
"category": {
"value": 1
}
},
{
"key": "ice_cider",
"doc_count": 1,
"category": {
"value": 1
}
},
{
"key": "jewelry_1",
"doc_count": 1,
"category": {
"value": 1
}
},
{
"key": "laces",
"doc_count": 1,
"category": {
"value": 1
}
},
{
"key": "stationery",
"doc_count": 1,
"category": {
"value": 1
}
}
]
}
}
}
答案 1 :(得分：0)
使用嵌套的elasticsearch Terms Aggregations与您目前使用品牌的方式相同，您可以生成如下结构的输出：
"aggregations": {
"groups": {
"buckets": [
{
"key": "xyz",
"categories_per_group": {
"buckets": [
{
"key": "shoes",
"doc_count": 2
},
{
"key": "t-shirt",
"doc_count": 2
}
]
}
},
{
"key": "abc",
"categories_per_group": {
"buckets": [
{
"key": "shoes",
"doc_count": 1
},
{
"key": "t-shirt",
"doc_count": 1
},
{
"key": "makeup",
"doc_count": 1
}
]
}
}
]
}
}
可能可以编写pipeline bucket script（ES 2.x）来收集群组存储区中不同类别的数量，如您所建议的那样。
但是，使用上面的聚合桶输出自己实现这种减少逻辑可能更简单，更快捷。