如何通过JSON将查询设置为Elasticsearch SearchRequest？

Question

Elasticsearch:6.1.2

我有一个通过JSON的输入查询,并希望使用高级Java API来使用该查询数据构建搜索请求.

String jsonQuery = "..."
SearchRequest searchRequest = new SearchRequest()
SearchSourceBuilder builder = ?
searchRequest.source(builder);

我尝试通过以下方式构建构建器:

SearchSourceBuilder.fromXContent(XContentType.JSON.xContent().createParser(NamedXContentRegistry.EMPTY, query));

但是产量:

引起:org.elasticsearch.ElasticsearchException:org.elasticsearch.common.xcontent.NamedXContentRegistry.parseNamedObject(NamedXContentRegistry.java:129)〜[elasticsearch-6.1.2.jar:6.1.2] at此解析器不支持namedObject org.elasticsearch.common.xcontent.support.AbstractXContentParser.namedObject(AbstractXContentParser.java:402)〜[elasticsearch-6.1.2.jar:6.1.2] at org.elasticsearch.index.query.AbstractQueryBuilder.parseInnerQueryBuilder(AbstractQueryBuilder.java) :313)〜[elasticsearch-6.1.2.jar:6.1.2]在org.elasticsearch.search.builder.SearchSourceBuilder.parseXContent(SearchSourceBuilder.java:1003)〜[elasticsearch-6.1.2.jar:6.1.2]在org.elasticsearch.search.builder.SearchSourceBuilder.fromXContent(SearchSourceBuilder.java:115)〜[elasticsearch-6.1.2.jar:6.1.2]

Answer 1

我现在以这种方式生成SearchSourceBuilder：

String query = "..."
SearchSourceBuilder searchSourceBuilder = new SearchSourceBuilder();
SearchModule searchModule = new SearchModule(Settings.EMPTY, false, Collections.emptyList());
try (XContentParser parser = XContentFactory.xContent(XContentType.JSON).createParser(new NamedXContentRegistry(searchModule
            .getNamedXContents()), query)) {
    searchSourceBuilder.parseXContent(parser);
}