API语法
聚合条件与query条件同级别,因此需要使用request.source()来指定聚合条件。
聚合的结果解析:
@Override public Map<String, List<String>> filters(RequestParams params) { try { // 1.准备Request SearchRequest request = new SearchRequest("hotel"); // 2.准备DSL // 2.1.query buildBasicQuery(params, request); // 2.2.设置size request.source().size(0); // 2.3.聚合 buildAggregation(request); // 3.发出请求 SearchResponse response = client.search(request, RequestOptions.DEFAULT); // 4.解析结果 Map<String, List<String>> result = new HashMap<>(); Aggregations aggregations = response.getAggregations(); // 4.1.根据品牌名称,获取品牌结果 List<String> brandList = getAggByName(aggregations, "brandAgg"); result.put("品牌", brandList); // 4.2.根据品牌名称,获取品牌结果 List<String> cityList = getAggByName(aggregations, "cityAgg"); result.put("城市", cityList); // 4.3.根据品牌名称,获取品牌结果 List<String> starList = getAggByName(aggregations, "starAgg"); result.put("星级", starList); return result; } catch (IOException e) { throw new RuntimeException(e); } } private void buildAggregation(SearchRequest request) { request.source().aggregation(AggregationBuilders .terms("brandAgg") .field("brand") .size(100) ); request.source().aggregation(AggregationBuilders .terms("cityAgg") .field("city") .size(100) ); request.source().aggregation(AggregationBuilders .terms("starAgg") .field("starName") .size(100) ); } private List<String> getAggByName(Aggregations aggregations, String aggName) { // 4.1.根据聚合名称获取聚合结果 Terms brandTerms = aggregations.get(aggName); // 4.2.获取buckets List<? extends Terms.Bucket> buckets = brandTerms.getBuckets(); // 4.3.遍历 List<String> brandList = new ArrayList<>(); for (Terms.Bucket bucket : buckets) { // 4.4.获取key String key = bucket.getKeyAsString(); brandList.add(key); } return brandList; }View Code
标签:聚合,实现,RestAPI,request,aggregations,source,result,List From: https://www.cnblogs.com/WarBlog/p/18676786