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进行的输入查询，并且想使用高级Java
API
使用该查询数据来构建搜索请求。

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

我试图通过以下方式构造构建器：

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

但这会产生：

由以下原因引起：org.elasticsearch.ElasticsearchException：org.elasticsearch.common.xcontent.NamedXContentRegistry.parseNamedObject（NamedXContentRegistry.java:129）〜[elasticsearch-6.1.2.jar：6.1.2]处的解析器不支持namedObject
org.elasticsearch.index.query.AbstractQueryBuilder.parseInnerQueryBuilder（AbstractQueryBuilder.java）上的org.elasticsearch.common.xcontent.support.AbstractXContentParser.namedObject（AbstractXContentParser.java:402）〜[elasticsearch-6.1.2.jar：6.1.2]
：313）〜[elasticsearch-6.1.2.jar：6.1.2]，位于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);
}