使用动态密钥将传入的JSON直接解析为DTO

Question

我正在尝试将Elastic搜索返回的数据解析为java类。 弹性搜索将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
           }
        }
     ]
   }
}

这里的“ aggregationName”和“ nestedAggregationName”是动态的，意味着它们可以是任何东西，那么我该如何管理我的课程呢？ 我想直接解析返回到DTO的数据。

编辑：我能够通过使类如下以一种通用的方式来解析“ aggregationName”：

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

     //with getters and setters.
}

因此，无论“ aggregationName”是什么，它都易于解析，并且AggregationModel包含与“ aggregationName”相对应的字段。 如何为“ nestedAggregationName”解决此问题？

请帮忙。

Answer 1

一段时间以前，我摸索了同样的问题，不幸的是Java客户端不支持聚合响应（如果我错了，请纠正我）。 以下是我经过多轮跟踪错误方法得出的结论，请告诉我是否有帮助。

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;
}

我知道这看起来有些笨拙，但可以为您服务。 否则让我知道。