我正在尝试对我拥有的一组数据执行一些弹性搜索查询。 我有一个用户文档，它是许多子页面视图文档的父级。我希望返回已经查看特定页面任意次数的所有用户（由用户输入框定义）。到目前为止，我有一个has_child查询，它将返回所有具有某些id的页面视图的用户。然而，这将使那些父母带着他们所有的孩子回归。接下来，我尝试在这些查询结果上编写聚合，这将基本上以聚合形式执行相同的has_child查询。现在，我有过滤子文档的正确文档计数。我需要使用此文档计数返回并过滤父项。要用单词解释查询，“将查看特定页面的所有用户返回给我4次以上”。我可能需要重构我的数据。有什么想法吗？
到目前为止，这是我的查询：
curl -XGET 'http://localhost:9200/development_users/_search?pretty=true' -d '
{
"query" : {
"has_child" : {
"type" : "page_view",
"query" : {
"terms" : {
"viewed_id" : [175,180]
}
}
}
},
"aggs" : {
"to_page_view": {
"children": {
"type" : "page_view"
},
"aggs" : {
"page_views_that_match" : {
"filter" : { "terms": { "viewed_id" : [175,180] } }
}
}
}
}
}'
这给我的回复如下：
{
"took" : 3,
"timed_out" : false,
"_shards" : {
"total" : 5,
"successful" : 5,
"failed" : 0
},
"hits" : {
"total" : 1,
"max_score" : 1.0,
"hits" : [ {
"_index" : "development_users",
"_type" : "user",
"_id" : "22548",
"_score" : 1.0,
"_source":{"id":22548,"account_id":1009}
} ]
},
"aggregations" : {
"to_page_view" : {
"doc_count" : 53,
"page_views_that_match" : {
"doc_count" : 2
}
}
}
}
相关映射：
{
"development_users" : {
"mappings" : {
"page_view" : {
"dynamic" : "false",
"_parent" : {
"type" : "user"
},
"_routing" : {
"required" : true
},
"properties" : {
"created_at" : {
"type" : "date",
"format" : "date_time"
},
"id" : {
"type" : "integer"
},
"viewed_id" : {
"type" : "integer"
},
"time_on_page" : {
"type" : "integer"
},
"title" : {
"type" : "string"
},
"type" : {
"type" : "string"
},
"updated_at" : {
"type" : "date",
"format" : "date_time"
},
"url" : {
"type" : "string"
}
}
},
"user" : {
"dynamic" : "false",
"properties" : {
"account_id" : {
"type" : "integer"
},
"id" : {
"type" : "integer"
}
}
}
}
}
}
答案 0 :(得分：5)
好的，所以这是一种参与。我做了一些简化，以保持在我脑海中。首先，我使用了这种映射：
PUT /test_index
{
"mappings": {
"page_view": {
"_parent": {
"type": "development_user"
},
"properties": {
"viewed_id": {
"type": "string"
}
}
},
"development_user": {
"properties": {
"id": {
"type": "string"
}
}
}
}
}
然后我添加了一些数据。在这个小小的宇宙中，我有三个用户和两个页面。我想找到至少两次查看"page_a"的用户，因此如果构建正确的查询，则只会返回用户3。
POST /test_index/development_user/_bulk
{"index":{"_type":"development_user","_id":1}}
{"id":"user_1"}
{"index":{"_type":"page_view","_parent":1}}
{"viewed_id":"page_a"}
{"index":{"_type":"development_user","_id":2}}
{"id":"user_2"}
{"index":{"_type":"page_view","_parent":2}}
{"viewed_id":"page_b"}
{"index":{"_type":"development_user","_id":3}}
{"id":"user_3"}
{"index":{"_type":"page_view","_parent":3}}
{"viewed_id":"page_a"}
{"index":{"_type":"page_view","_parent":3}}
{"viewed_id":"page_a"}
{"index":{"_type":"page_view","_parent":3}}
{"viewed_id":"page_b"}
要获得该答案，我们将使用聚合。请注意，我不想要返回文档（正常方式），但我确实希望过滤掉我们分析的文档，因为它会提高效率。所以我使用的是与之前相同的基本过滤器。
因此聚合树以terms_parent_id开头，它只是分隔父文档。在里面，我children_page_view将子文档过滤到我想要的文件（"page_a"），并且层次结构旁边的bucket_selector_page_id_term_count使用bucket selector（您需要ES 2.x）来满足那些符合标准的人的父文件，然后最后一个top hits aggregation向我们展示符合要求的文件。
POST /test_index/development_user/_search
{
"size": 0,
"query": {
"has_child": {
"type": "page_view",
"query": {
"terms": {
"viewed_id": [
"page_a"
]
}
}
}
},
"aggs": {
"terms_parent_id": {
"terms": {
"field": "id"
},
"aggs": {
"children_page_view": {
"children": {
"type": "page_view"
},
"aggs": {
"filter_page_ids": {
"filter": {
"terms": {
"viewed_id": [
"page_a"
]
}
}
}
}
},
"bucket_selector_page_id_term_count": {
"bucket_selector": {
"buckets_path": {
"children_count": "children_page_view>filter_page_ids._count"
},
"script": "children_count >= 2"
}
},
"top_hits_users": {
"top_hits": {
"_source": {
"include": [
"id"
]
}
}
}
}
}
}
}
返回：
{
"took": 14,
"timed_out": false,
"_shards": {
"total": 5,
"successful": 5,
"failed": 0
},
"hits": {
"total": 2,
"max_score": 0,
"hits": []
},
"aggregations": {
"terms_parent_id": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "user_3",
"doc_count": 1,
"children_page_view": {
"doc_count": 3,
"filter_page_ids": {
"doc_count": 2
}
},
"top_hits_users": {
"hits": {
"total": 1,
"max_score": 1,
"hits": [
{
"_index": "test_index",
"_type": "development_user",
"_id": "3",
"_score": 1,
"_source": {
"id": "user_3"
}
}
]
}
}
}
]
}
}
}
这是我使用的所有代码：
http://sense.qbox.io/gist/43f24461448519dc884039db40ebd8e2f5b7304f