弹性搜索索引中有以下文档。
#include <iostream>
#include <fstream>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
int main(){
vector <int> nums;
string str;
cin >> str;
int lastcomma = -1;
while(str.find(',', lastcomma+1) != string::npos){ // find the next comma
int curr = str.find(',', lastcomma+1);
// stoi converts a string to an integer; just what you need
nums.push_back(stoi(str.substr(lastcomma+1, curr - (lastcomma+1))));
lastcomma = curr;
}
// get the last number
nums.push_back(stoi(str.substr(lastcomma+1, str.size()-(lastcomma+1))));
return 0;
}
我要从中将结果[]留空。在这里我们可以看到两个文档的uid相同。我正在使用以下查询来获取结果：
[{
"_index": "ten2",
"_type": "documents",
"_id": "c323c2244a4a4c22_en-us",
"_source": {
"publish_details": [{
"environment": "603fe91adbdcff66",
"time": "2020-06-24T13:36:55.514Z",
"locale": "hi-in",
"user": "aadab2f531206e9d",
"version": 1
},
{
"environment": "603fe91adbdcff66",
"time": "2020-06-24T13:36:55.514Z",
"locale": "en-us",
"user": "aadab2f531206e9d",
"version": 1
}
],
"created_at": "2020-06-24T13:36:43.037Z",
"_in_progress": false,
"title": "Entry 1",
"locale": "en-us",
"url": "/entry-1",
"tags": [],
"uid": "c323c2244a4a4c22",
"updated_at": "2020-06-24T13:36:43.037Z",
"fields": []
}
},
{
"_index": "ten2",
"_type": "documents",
"_id": "c323c2244a4a4c22_mr-in",
"_source": {
"publish_details": [{
"environment": "603fe91adbdcff66",
"time": "2020-06-24T13:37:26.205Z",
"locale": "mr-in",
"user": "aadab2f531206e9d",
"version": 1
}],
"created_at": "2020-06-24T13:36:43.037Z",
"_in_progress": false,
"title": "Entry 1 marathi",
"locale": "mr-in",
"url": "/entry-1",
"tags": [],
"uid": "c323c2244a4a4c22",
"updated_at": "2020-06-24T13:37:20.092Z",
"fields": []
}
}
]
但是上面的查询给了我全部2个文档，但是我希望得到结果作为银行，这里的原因是uid很常见，并且uid包含发布详细信息中的所有三个local。因此，获取有效结果的方法是，是否有任何聚合查询在这里对我有帮助。这只是一个示例，我有很多文档要过滤掉。 Kindle在这里帮助我。
答案 0 :(得分：1)
{
"aggs": {
"agg1": {
"terms": {
"field": "uid.raw"
},
"aggs": {
"agg2": {
"nested": {
"path": "publish_details"
},
"aggs": {
"locales": {
"terms": {
"field": "publish_details.locale"
}
}
}
}
}
}
}
}
此查询将先按uid然后按publish_details.locale分组您
它提供如下结果
"aggregations": {
"agg1": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "c323c2244a4a4c22",
"doc_count": 2,
"agg2": {
"doc_count": 3,
"locales": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "en-us",
"doc_count": 1
},
{
"key": "hi-in",
"doc_count": 1
},
{
"key": "mr-in",
"doc_count": 1
}
]
}
}
},
{
"key": "c323c2244rrffa4a4c22",
"doc_count": 1,
"agg2": {
"doc_count": 2,
"locales": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "en-us",
"doc_count": 1
},
{
"key": "hi-in",
"doc_count": 1
}
]
}
}
}
]
我有3个文档，其中两个具有相同的ID，而另一个则不同。
我将进一步更新查询，以删除您有3个存储桶的第一个结果。您也可以在代码中进一步处理它。
您可以做到。 10k个文档很好。但是当您拥有数以百万计的资产时，您应该有足够的资源来执行此操作。
{
"size" : 0,
"query":{
"bool" :{
"must_not":{
"match":{
"publish_details.environment":"603fe91adbdcff66"
}
}
}
},
"aggs": {
"uids": {
"terms": {
"field": "uid.raw"
},
"aggs": {
"details": {
"nested": {
"path": "publish_details"
},
"aggs": {
"locales": {
"terms": {
"field": "publish_details.locale"
}
},
"unique_locales": {
"value_count": {
"field": "publish_details.locale"
}
}
}
}
}
}
}
}
结果：
"aggregations": {
"uids": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "c323c2244a4a4c22",
"doc_count": 2,
"details": {
"doc_count": 3,
"locales": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "en-us",
"doc_count": 1
},
{
"key": "hi-in",
"doc_count": 1
},
{
"key": "mr-in",
"doc_count": 1
}
]
},
"unique_locales": {
"value": 3
}
}
},
{
"key": "c323c2244rrffa4a4c22",
"doc_count": 1,
"details": {
"doc_count": 2,
"locales": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "en-us",
"doc_count": 1
},
{
"key": "hi-in",
"doc_count": 1
}
]
},
"unique_locales": {
"value": 2
}
}
}
]