在post索引中，postid是主键，userid是外键。
我想要所有的帖子，但只从一个用户ID发布，这样只有一个用户在结果中有一个帖子，按postdate排序(可选的最新优先)
//Actual Result
[
{
userid: "u1",
postid: "p1"
},
{
userid: "u1",
postid: "p2"
},
{
userid: "u2",
postid: "p3"
},
{
userid: "u3",
postid: "p4"
},
{
userid: "u3",
postid: "p5"
},
{
userid: "u3",
postid: "p6"
}
]
以下所需
//Expecting Result
[
{
userid: "u1",
postid: "p1"
},
{
userid: "u2",
postid: "p3"
},
{
userid: "u3",
postid: "p4"
}
]
我认为您可以使用top hit。这里是这个的例子：
DELETE my-index-000001
PUT my-index-000001
{
"mappings": {
"properties": {
"userid": {
"type": "keyword"
},
"postid": {
"type": "keyword"
},
"postdate": {
"type": "date"
}
}
}
}
PUT my-index-000001/_doc/1
{"userid": "u1", "postid": "p1", "postdate": "2021-03-01"}
PUT my-index-000001/_doc/2
{"userid": "u1", "postid": "p2", "postdate": "2021-03-02"}
PUT my-index-000001/_doc/3
{"userid": "u2", "postid": "p3", "postdate": "2021-03-03"}
PUT my-index-000001/_doc/4
{"userid": "u3", "postid": "p4", "postdate": "2021-03-04"}
PUT my-index-000001/_doc/5
{"userid": "u3", "postid": "p5", "postdate": "2021-03-05"}
PUT my-index-000001/_doc/6
{"userid": "u3", "postid": "p6", "postdate": "2021-03-06"}
以下是创建索引的示例步骤。这里的查询：
GET my-index-000001/_search
{
"size": 0,
"aggs": {
"top_users": {
"terms": {
"field": "userid",
"size": 100
},
"aggs": {
"top": {
"top_hits": {
"sort": [
{
"postdate": {
"order": "desc"
}
}
],
"_source": {
"includes": [ "postdate", "postid" ]
},
"size": 1
}
}
}
}
}
}
在结果集中，您可以看到聚合中每个用户的顶部帖子：
{
"took" : 3,
"timed_out" : false,
"_shards" : {
"total" : 1,
"successful" : 1,
"skipped" : 0,
"failed" : 0
},
"hits" : {
"total" : {
"value" : 6,
"relation" : "eq"
},
"max_score" : null,
"hits" : [ ]
},
"aggregations" : {
"top_users" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : "u3",
"doc_count" : 3,
"top" : {
"hits" : {
"total" : {
"value" : 3,
"relation" : "eq"
},
"max_score" : null,
"hits" : [
{
"_index" : "my-index-000001",
"_type" : "_doc",
"_id" : "6",
"_score" : null,
"_source" : {
"postdate" : "2021-03-06",
"postid" : "p6"
},
"sort" : [
1614988800000
]
}
]
}
}
},
{
"key" : "u1",
"doc_count" : 2,
"top" : {
"hits" : {
"total" : {
"value" : 2,
"relation" : "eq"
},
"max_score" : null,
"hits" : [
{
"_index" : "my-index-000001",
"_type" : "_doc",
"_id" : "2",
"_score" : null,
"_source" : {
"postdate" : "2021-03-02",
"postid" : "p2"
},
"sort" : [
1614643200000
]
}
]
}
}
},
{
"key" : "u2",
"doc_count" : 1,
"top" : {
"hits" : {
"total" : {
"value" : 1,
"relation" : "eq"
},
"max_score" : null,
"hits" : [
{
"_index" : "my-index-000001",
"_type" : "_doc",
"_id" : "3",
"_score" : null,
"_source" : {
"postdate" : "2021-03-03",
"postid" : "p3"
},
"sort" : [
1614729600000
]
}
]
}
}
}
]
}
}
}
假设以下形式的索引映射：
PUT user_posts
{
"mappings": {
"properties": {
"userid": {
"type": "keyword"
},
"postid": {
"type": "keyword"
},
"postdate": {
"type": "date"
}
}
}
}
你可以：
- 在
userid上聚合，并按字母顺序排列ID - 2上的子聚合，并通过CCD_ 4聚合按CCD_
- 通过
filter_path选项筛选响应以仅检索您需要的内容
POST user_posts/_search?filter_path=aggregations.*.buckets.key,aggregations.*.buckets.*.buckets.key
{
"size": 0,
"aggs": {
"by_userid": {
"terms": {
"field": "userid",
"order": {
"_key": "asc"
},
"size": 100
},
"aggs": {
"by_latest_postid": {
"terms": {
"field": "postid",
"size": 1,
"order": {
"latest_posttime": "desc"
}
},
"aggs": {
"latest_posttime": {
"max": {
"field": "postdate"
}
}
}
}
}
}
}
}
收益率：
{
"aggregations" : {
"by_userid" : {
"buckets" : [
{
"key" : "u1",
"by_latest_postid" : {
"buckets" : [
{
"key" : "p1"
}
]
}
},
{
"key" : "u2",
"by_latest_postid" : {
"buckets" : [
{
"key" : "p3"
}
]
}
},
{
"key" : "u3",
"by_latest_postid" : {
"buckets" : [
{
"key" : "p4"
}
]
}
}
]
}
}
}
然后你可以像往常一样发布流程：
...
const response = await ...; // transform the above request for use in the ES JS lib of your choice
const result = response.aggregations.by_userid.buckets.map(b => {
return {
userid: b.key,
postid: b.by_latest_postid.buckets && b.by_latest_postid.buckets[0].key
}
})
您可以使用热门子聚合。因此，首先通过userId进行terms聚合，然后您可以使用按post-date排序的热门文章来获得每个用户的最新帖子。
我应该说，如果你有很多userId，并且你想要每个CCD_9的最高命中率，你可能应该使用复合聚合作为你的顶级agg，而不是术语。