如何使用jaxb将xml字符串解组为java对象

Question

我需要使用rest模板发送请求。在发送请求之前，我需要在发送请求时将对象整理成xml。我收到了请求的响应，但是XML格式的。然后我需要将响应xml转换为对象，以便在接口上显示结果

下面是我发送请求的控制器

 @RequestMapping("/searchSummon")
 public String Search(Model model)
 {

    model.addAttribute("jaxbExample", new JAXBExample());
    model.addAttribute("pdxiRes", new PDXIRes());

    JAXBExample jaxbExample = new JAXBExample();
    String create_xml = jaxbExample.CreateXML();

    System.out.println(create_xml);

    RestTemplate restTemplate = new RestTemplate();
    String a = restTemplate.postForObject("http://192.168.80.30/summon-
    V2/example", "<?xml version='1.0' encoding='UTF-8'?> <!DOCTYPE PDXIReq 
    SYSTEM 'summon.dtd'>" + create_xml,String.class);

    System.out.println(a);

    return "searchSummon";
}

我怎样才能将“a”解组到对象？ 课堂反应 PDXIRes 标题 要求 细节 状态

用于响应（'a'）的xml

 <?xml version="1.0" encoding="utf-8"?>
 <PDXIRes>
 <header>
    <sp_code>abc017637m</sp_code>
 </header>
 <request id="1sq1216272728732">
    <id_no>683642435</id_no>
    <name>SALLY</name>
    <max_index>1024</max_index>
    <total_summons>2</total_summons>
    <summons_detail>
        <row num="1">
            <summons_id>1810000200002AQ639332</summons_id>
            <vehicle>NN162</vehicle>

        </row>
        <row num="2">
            <summons_id>1810000200002AM947772</summons_id>
            <vehicle>NN162</vehicle
        </row>
    </summons_detail>
    <status>
        <status_code>01</status_code>
        <status_msg>Successful</status_msg>
    </status>
   </request>
   </PDXIRes>

Answer 1

可以使用解组器来实现这一点

JAXBContext jaxbContext = JAXBContext.newInstance(Person.class);
Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();

StringReader reader = new StringReader("xml string here");
Person person = (Person) unmarshaller.unmarshal(reader);

在这里找到：Use JAXB to create Object from XML String