Question

我有来自elasticsearch rest客户端的json响应。我想从该字符串（json）创建elasticsearch SearchResponse或GetResponse对象，以便我可以重用grails-2.4.3 elasticsearch插件中的un-marshaling部分。有人可以帮助我吗？

Answer 1

我不确定这个问题是否仍然相关，但这个问题对我有用：

String responseJson = "{\"took\":5,\"timed_out\":false,\"_shards\".....}";
try {
  JsonXContentParser xContentParser = new JsonXContentParser(NamedXContentRegistry.EMPTY,
  new JsonFactory().createParser(responseJson));      
  SearchResponse response = SearchResponse.fromXContent(xContentParser);
  ... 
  Do Whatever
  ...
} catch (IOException e) {
  handleException....
}

Answer 2

我确实找到了可以帮助您的东西。

我写了这样的JSON：

XContentBuilder builder = XContentFactory.jsonBuilder();
response.toXContent(builder, ToXContent.EMPTY_PARAMS);
String result = Strings.toString(builder);

然后我想像这样阅读它：

 try {
     NamedXContentRegistry registry = new NamedXContentRegistry(getDefaultNamedXContents());
     XContentParser parser = JsonXContent.jsonXContent.createParser(registry, DeprecationHandler.THROW_UNSUPPORTED_OPERATION, result);
     SearchResponse searchResponse = SearchResponse.fromXContent(parser);
 } catch (IOException e) {
     System.out.println("exception " + e);
 } catch (Exception e) {
     System.out.println("exception " + e);
 }

public static List<NamedXContentRegistry.Entry> getDefaultNamedXContents() {
    Map<String, ContextParser<Object, ? extends Aggregation>> map = new HashMap<>();
    map.put(TopHitsAggregationBuilder.NAME, (p, c) -> ParsedTopHits.fromXContent(p, (String) c));
    map.put(StringTerms.NAME, (p, c) -> ParsedStringTerms.fromXContent(p, (String) c));
    List<NamedXContentRegistry.Entry> entries = map.entrySet().stream()
            .map(entry -> new NamedXContentRegistry.Entry(Aggregation.class, new ParseField(entry.getKey()), entry.getValue()))
            .collect(Collectors.toList());
    return entries;
}

希望它能起作用：）

字符串json到elasticSearch SearchResponse对象？

2 个答案: