首页 \ 问答 \ 以json格式返回solr响应(Return solr response in json format)

以json格式返回solr响应(Return solr response in json format)

我试图以JSON格式返回solr响应。 但是我无法获得JSON响应。 下面是我的solr配置:'

<queryResponseWriter name="json" default="true" class="org.apache.solr.request.JSONResponseWriter">
    <str name="content-type">application/json</str>
</queryResponseWriter>

下面是我的java代码:

HttpSolrServer server = new HttpSolrServer(serverUrl);
SolrQuery query = new SolrQuery();
query.setQuery(query);
query.set("indent","true");
query.set("wt","json");
QueryResponse response = server.query(query);

System.out.println(response.getResults().get(index));

但是输出以下列格式显示。

{numFound=1,start=0,docs=[SolrDocument{_uniqueKey=[“abc@gmail.com”,”abc”], calculation_policy=text_value, username=abc, email=abc@gmail.com, display_order=10, last_login=Mon Jan 26 11:27:35 PST 2015, created=Mon Jan 26 11:27:35 PST 2015}]}

但是,执行以下行后,我得到了JSON响应:

System.out.println(new Gson().toJson(queryResponse.getResults().get(index)));

有人可以告诉我,我错过了哪一步?


I am trying to return solr response in JSON format. However I am not able to get the JSON response. Below is my solr config:'

<queryResponseWriter name="json" default="true" class="org.apache.solr.request.JSONResponseWriter">
    <str name="content-type">application/json</str>
</queryResponseWriter>

Below is my java code:

HttpSolrServer server = new HttpSolrServer(serverUrl);
SolrQuery query = new SolrQuery();
query.setQuery(query);
query.set("indent","true");
query.set("wt","json");
QueryResponse response = server.query(query);

System.out.println(response.getResults().get(index));

However output is displyed in following format.

{numFound=1,start=0,docs=[SolrDocument{_uniqueKey=[“abc@gmail.com”,”abc”], calculation_policy=text_value, username=abc, email=abc@gmail.com, display_order=10, last_login=Mon Jan 26 11:27:35 PST 2015, created=Mon Jan 26 11:27:35 PST 2015}]}

However I get JSON response after execute following line:

System.out.println(new Gson().toJson(queryResponse.getResults().get(index)));

Can someone please tell me what step am I missing?


原文:https://stackoverflow.com/questions/28374428
更新时间:2022-03-18 18:03

最满意答案

您没有使用泛型(您使用的是原始类型 ),因此编译器不知道映射中的值是什么类型 - map.get(k)返回Object ,就编译器而言, no +(Object, int)运算符。

只需在map声明和初始化中使用泛型:

// Java 7 and later, using the "diamond operator"
Map<Integer, Integer> map = new HashMap<>();

要么:

// Java 5 and 6 (will work with 7 and later too, but is longer)
Map<Integer, Integer> map = new HashMap<Integer, Integer>();

其余的将编译没有问题。

如果您是泛型新手 ,请参阅教程综合FAQ 。 你应该很少使用原始类型(几乎不会)。 你应该能够配置你的IDE或者编译器给你一个警告,如果你这样做的话(例如用-Xlint for javac )。


You're not using generics (you're using a raw type instead), so the compiler has no idea what type of value is in the map - map.get(k) returns Object as far as the compiler is concerned, and there's no +(Object, int) operator.

Just use generics in your map declaration and initialization:

// Java 7 and later, using the "diamond operator"
Map<Integer, Integer> map = new HashMap<>();

Or:

// Java 5 and 6 (will work with 7 and later too, but is longer)
Map<Integer, Integer> map = new HashMap<Integer, Integer>();

The rest will compile with no problems.

If you're new to generics, see the tutorial and the comprehensive FAQ. You should very, very rarely use raw types (almost never). You should be able to configure your IDE or compiler to give you a warning if you do so (e.g. with -Xlint for javac).

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