使用Solr Suggester从solrj获取结果的正确方法是什么?
这是我的要求:
SolrQuery query = new SolrQuery();
query.setRequestHandler("/suggest");
query.setParam("suggest", "true");
query.setParam("suggest.build", "true");
query.setParam("suggest.dictionary", "mySuggester");
query.setParam("suggest.q", "So");
QueryResponse response = server.query(query);
但是我发现得到回应非常困难。我得到回应的方式是:
NamedList obj = (NamedList)((Map)response.getResponse().get("suggest")).get("mySuggester");
SimpleOrderedMap obj2 = (SimpleOrderedMap) obj.get("So");
List<SimpleOrderedMap> obj3 = (List<SimpleOrderedMap>) obj2.get("suggestions");
这似乎假设了很多关于我从响应中获得的对象,并且很难预测错误。
有没有比这更好更清洁的方式?
答案 0 :(得分:4)
在新版本中有一个SuggesterResponse:
答案 1 :(得分:1)
最好的选择是将它作为List,下面代码为我工作
HttpSolrClient solrClient = new HttpSolrClient(solrURL);
SolrQuery query = new SolrQuery();
query.setRequestHandler("/suggest");
query.setParam("suggest.q", "Ins");
query.setParam("wt", "json");
try {
QueryResponse response = solrClient.query(query);
System.out.println(response.getSuggesterResponse().getSuggestedTerms());
List<String> types=response.getSuggesterResponse().getSuggestedTerms().get("infixSuggester");
System.out.println(types);
} catch (SolrServerException | IOException e) {
e.printStackTrace();
}
答案 2 :(得分:0)
您可以通过SpellCheckResponse
通过以下方式获取建议
SpellCheckResponse spellCheckResponse=response.getSpellCheckResponse();
查看此link了解详情