ES High Level REST Client API 查询 聚合
1 准备数据1.1 插入测试数据2 Maven引入ES3 创建Client4 查询API4.1 根据id查询单条数据4.2 根据多个id查询4.3 根据条件分页查询4.4 count4.5 max5 terms聚合5.1 根据1个字段group by5.2 根据多个字段group by关于聚合的博客：
Elasticsearch教程(3) ES聚合查询DSL
Elasticsearch教程(4) High Level REST Client API 查询 聚合 分组
Elasticsearch教程(5) 指标聚合 SQL DSL JavaAPI
Elasticsearch教程(6) 桶聚合Query DSL-Terms Aggregation
Elasticsearch教程(10) ES term terms prefix 搜索 聚合查询 详细总结
Elasticsearch教程(11) elasticsearch 桶聚合 Query DSL
Elasticsearch教程(32) ES 聚合查询后过滤 Distinct Group By Having功能
1 准备数据
首先启动好的ES6.8服务和Kibana服务
1.1 插入测试数据
假设有个部门组织的场景，部门有上级部门或者下级部门，像一颗树那样
PUT /dept/_doc/1{"code": "dept_1","name": "部门1","level": 1,"path": "1","parentId": "","status":1}PUT /dept/_doc/2{"code": "dept_2","name": "部门2","level": 1,"path": "2","parentId": "","status":0}PUT /dept/_doc/3{"code": "dept_1_1","name": "部门1_1","level": 2,"path": "1,3","parentId": "1","status":0}PUT /dept/_doc/4{"code": "dept_1_2","name": "部门1_2","level": 2,"path": "1,4","parentId": "1","status":0}PUT /dept/_doc/5{"code": "dept_1_1_1","name": "部门1_1_1","level": 3,"path": "1,3,5","parentId": "3","status":1}PUT /dept/_doc/6{"code": "dept_1_1_2","name": "部门1_1_2","level": 3,"path": "1,3,6","parentId": "3","status":null}
数据结果如下：
2 Maven引入ES
新建一个空的Maven结构项目，在pom.xml里添加如下
<dependency><groupId>org.elasticsearch.client</groupId><artifactId>elasticsearch-rest-high-level-client</artifactId><version>6.8.4</version></dependency>
3 创建Client
public class EsRestUtils {private static RestHighLevelClient client;private static final String type = "_doc";public static RestHighLevelClient getClient() {if (client == null) {client = new RestHighLevelClient(RestClient.builder(new HttpHost("192.168.8.201", 9200, "http")));}return client;}}
4 查询API
4.1 根据id查询单条数据
用SQL描述就是
select * from dept where id = '1'
protected static Map<String, Object> getById(String index, String id) throws IOException {getClient();GetRequest getRequest = new GetRequest(index, type, id);GetResponse getResponse = client.get(getRequest, RequestOptions.DEFAULT);if (getResponse.isExists()){return getResponse.getSourceAsMap();}return null;}
4.2 根据多个id查询
用SQL描述就是
select * from dept where id in ("1", "2", "3")
protected static List<Map<String, Object>> getByIds(String index, List<String> ids) throws IOException {getClient();List<Map<String, Object>> results = new ArrayList<>();MultiGetRequest request = new MultiGetRequest();ids.stream().forEach(id -> {request.add(new MultiGetRequest.Item(index, type, id));});MultiGetResponse response = client.mget(request, RequestOptions.DEFAULT);GetResponse getResponse;for (int i = 0; i < response.getResponses().length; i++) {getResponse = response.getResponses()[i].getResponse();if (getResponse.isExists()) {results.add(getResponse.getSourceAsMap());}}return results;}
4.3 根据条件分页查询
用SQL描述就是
select * from dept where id in ("1", "2", "3") limt 2,10
protected static List<Map<String, Object>> getByWhere(String index, QueryBuilder queryBuilder, int pageNo, int pageSize) throws IOException {getClient();List<Map<String, Object>> results = new ArrayList<>();SearchSourceBuilder searchSourceBuilder = new SearchSourceBuilder();searchSourceBuilder.query(queryBuilder);searchSourceBuilder.from(pageNo);searchSourceBuilder.size(pageSize);SearchRequest searchRequest = new SearchRequest(index).types(type);searchRequest.source(searchSourceBuilder);SearchResponse searchResponse = client.search(searchRequest, RequestOptions.DEFAULT);SearchHits hits = searchResponse.getHits();SearchHit[] searchHits = hits.getHits();for (SearchHit hit : searchHits) {results.add(hit.getSourceAsMap());}return results;}
4.4 count
用SQL描述就是
select count(1) from dept where name like '部门%'
public static long count(QueryBuilder queryBuilder, String... indexs) throws IOException {getClient();SearchSourceBuilder searchSourceBuilder = new SearchSourceBuilder();searchSourceBuilder.query(queryBuilder);CountRequest countRequest = new CountRequest(indexs);countRequest.source(searchSourceBuilder);CountResponse countResponse = client.count(countRequest, RequestOptions.DEFAULT);long count = countResponse.getCount();return count;}
4.5 max
用SQL描述就是
select max(level) from dept where name like '部门%'
public static Double getMax(QueryBuilder queryBuilder, String field, String... indexs) throws IOException {getClient();SearchSourceBuilder searchSourceBuilder = new SearchSourceBuilder();searchSourceBuilder.query(queryBuilder);searchSourceBuilder.size(0);AggregationBuilder aggregationBuilder = AggregationBuilders.max("agg").field(field);searchSourceBuilder.aggregation(aggregationBuilder);SearchRequest searchRequest = new SearchRequest(indexs).types(type);searchRequest.source(searchSourceBuilder);SearchResponse searchResponse = client.search(searchRequest, RequestOptions.DEFAULT);Max agg = searchResponse.getAggregations().get("agg");return agg.getValue();}
5 terms聚合
5.1 根据1个字段group by
用SQL描述就是
select level, count(id) from dept where name like '部门%' group by level
public static Map<String, Long> getTermsAgg(QueryBuilder queryBuilder, String field, String... Map<String, Long> groupMap = new HashMap<>();getClient();SearchSourceBuilder searchSourceBuilder = new SearchSourceBuilder();searchSourceBuilder.query(queryBuilder);searchSourceBuilder.size(0);AggregationBuilder aggregationBuilder = AggregationBuilders.terms("agg").field(field);searchSourceBuilder.aggregation(aggregationBuilder);SearchRequest searchRequest = new SearchRequest(indexs).types(type);searchRequest.source(searchSourceBuilder);SearchResponse searchResponse = client.search(searchRequest, RequestOptions.DEFAULT);Terms terms = searchResponse.getAggregations().get("agg");for (Terms.Bucket entry : terms.getBuckets()) {groupMap.put(entry.getKey().toString(), entry.getDocCount());}return groupMap;}
写代码测试
protected static void testGetTermsAgg(String index) {QueryBuilder queryBuilder = QueryBuilders.wildcardQuery("name.keyword", "部门*");try {Map<String, Long> groupMap = EsRestUtils.getTermsAgg(queryBuilder, "level", index);groupMap.forEach((key, value) -> System.out.println(key + " -> " + value.toString()));} catch (IOException e) {e.printStackTrace();}}
运行结果如下，左边是level，右边是个数
1 -> 22 -> 23 -> 2
5.2 根据多个字段group by
用SQL描述就是
select level, status, count(id) from dept where name like '部门%' group by level, status
public static Map<String, Map<String, Long>> getTermsAggTwoLevel(QueryBuilder queryBuilder, String field1, String field2, String... indexs) throws IOException {Map<String, Map<String, Long>> groupMap = new HashMap<>();getClient();SearchSourceBuilder searchSourceBuilder = new SearchSourceBuilder();searchSourceBuilder.query(queryBuilder);searchSourceBuilder.size(0);AggregationBuilder agg1 = AggregationBuilders.terms("agg1").field(field1);AggregationBuilder agg2 = AggregationBuilders.terms("agg2").field(field2);agg1.subAggregation(agg2);searchSourceBuilder.aggregation(agg1);SearchRequest searchRequest = new SearchRequest(indexs).types(type);searchRequest.source(searchSourceBuilder);SearchResponse searchResponse = client.search(searchRequest, RequestOptions.DEFAULT);Terms terms1 = searchResponse.getAggregations().get("agg1");Terms terms2;for (Terms.Bucket bucket1 : terms1.getBuckets()) {terms2 = bucket1.getAggregations().get("agg2");Map<String, Long> map2 = new HashMap<>();for (Terms.Bucket bucket2 : terms2.getBuckets()) {map2.put(bucket2.getKey().toString(), bucket2.getDocCount());}groupMap.put(bucket1.getKey().toString(), map2);}return groupMap;}
写代码测试
protected static void testGetTermsAgg2(String index) {QueryBuilder queryBuilder = QueryBuilders.wildcardQuery("name.keyword", "部门*");try {Map<String, Map<String, Long>> groupMap = EsRestUtils.getTermsAggTwoLevel(queryBuilder, "level", "status", index);groupMap.forEach((key, value) -> System.out.println(key + " -> " + value.toString()));} catch (IOException e) {e.printStackTrace();}}
对于id=6的那条数据，status=null，就不会统计到，如果没有status也不会统计到。
1 -> {0=1, 1=1}#leve=1的数据中，status=1的1条，status=0的1条2 -> {0=2}#leve=2的数据中，status=0的2条3 -> {1=1}#leve=3的数据中，status=1的1条
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