Question

我正在尝试将从Elastic搜索返回的数据解析为java类。弹性搜索以下面给出的形式返回数据：

"aggregations": {
  "aggregationName": {
     "doc_count_error_upper_bound": 0,
     "sum_other_doc_count": 0,
     "buckets": [
        {
           "key": 439,
           "doc_count": 7,
           "nestedAggregationName": {
              "value": 5
           }
        }
     ]
   }
}

这里＆＃34; aggregationName＆＃34;和＆＃34; nestedAggregationName＆＃34;名字是动态的，意味着它们可以是任何东西，所以我如何管理这是我的班级？我想直接解析返回到DTO的数据。

编辑：我能够解析＆＃34; aggregationName＆＃34;通过下面的课程以通用方式：

public Class Aggregation{
     private Map<String, AggregationModel> aggregation;

     //with getters and setters.
}

无论＆＃34; aggregationName＆＃34;是，它被轻松解析，AggregationModel包含对应于＆＃34; aggregationName＆＃34;的字段。如何为＆＃34; nestedAggregationName＆＃34; 解决此问题？

请帮忙。

Answer 1

我在一段时间后偶然发现了同样的问题，不幸的是java客户端不支持聚合响应（如果我错了请纠正我）。以下是我在多轮追踪错误方法后想出的，如果有帮助，请告诉我。

private Map<String, List<Map.Entry<String, Long>>> buildAggregationsFromResult(SearchResults resultSet, Set aggregationsName) {
    Map<String, List<Map.Entry<String, Long>>> aggregations = new LinkedHashMap<String, List<Map.Entry<String, Long>>>();
    if (null != resultSet.getAggregations()) {
        Map<String, Aggregation> aggregationMap = resultSet.getAggregations().asMap();
        for (Map.Entry<String, Aggregation> aggregationEntry : aggregationMap.entrySet()) {
            fetchBucketTermsFromAggregations(aggregationEntry,aggregations, aggregationsName);
        }
    }
    return aggregations;
}

private List<Map.Entry<String, Long>> fetchBucketTermsFromAggregations(Map.Entry<String, Aggregation> aggregationEntry, Map<String, List<Map.Entry<String, Long>>> aggregations, Set<String> aggregationsName) {
    Terms terms = null;
    List<Map.Entry<String, Long>> pairs = new ArrayList<>();
    if (aggregationEntry.getValue() instanceof InternalNested) {
        for (Map.Entry<String, Aggregation> aggregationEntryNested : ((InternalNested) aggregationEntry.getValue()).getAggregations().asMap().entrySet()) {
            pairs = fetchBucketTermsFromAggregations(aggregationEntryNested, aggregations, aggregationsName);
            if (CollectionUtils.isNotEmpty(pairs) && aggregationsName.contains(aggregationEntry.getKey()))
                aggregations.put(aggregationEntry.getKey(), pairs);

        }
    } else if (aggregationEntry.getValue() instanceof InternalFilters) {
        for (InternalFilters.Bucket buckets : ((InternalFilters) aggregationEntry.getValue()).getBuckets()) {
            for (Map.Entry<String, Aggregation> nestedAggregations : buckets.getAggregations().asMap().entrySet()) {
                pairs = fetchBucketTermsFromAggregations(nestedAggregations, aggregations, aggregationsName);
                if (CollectionUtils.isNotEmpty(pairs) && aggregationsName.contains(aggregationEntry.getKey()) )
                    aggregations.put(aggregationEntry.getKey(), pairs);
            }
        }
    } else {
        terms = (Terms) aggregationEntry.getValue();
        if (null != terms ) {
            Collection<Terms.Bucket> buckets = terms.getBuckets();
            for (Terms.Bucket bucket : buckets) {
                if (CollectionUtils.isNotEmpty(bucket.getAggregations().asList()))
                    pairs.add(new AbstractMap.SimpleEntry<String, Long>(bucket.getKeyAsString(), ((InternalReverseNested)bucket.getAggregations().asList().get(0)).getDocCount()));
                else
                    pairs.add(new AbstractMap.SimpleEntry<String, Long>(bucket.getKeyAsString(), bucket.getDocCount()));
            }
        }

        if (CollectionUtils.isNotEmpty(pairs) && aggregationsName.contains(aggregationEntry.getKey()))
            aggregations.put(aggregationEntry.getKey(), pairs);
    }
    return pairs;
}

我知道这看起来有点笨拙，但可能适合你的情况。别的我知道了。

使用动态密钥将传入的JSON直接解析到DTO

1 个答案: