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我需要使用rest模板发送请求。在发送之前,我需要在发送请求时将对象编组为xml。我从请求中获得了响应,但采用了XML格式。然后,我需要将响应xml转换为对象,以便将结果显示在界面上。如何使用jaxb解组xml字符串到java对象
下面是我的控制器,其中i发送请求
@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";
}
我怎么能解组 'a' 到对象? 类响应 PDXIRes 头 请求 详细 状态
XML的响应( 'A')
<?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>
如果你想使用REST框架类似新泽西编组/解组过程是自动处理... –
请检查https://stackoverflow.com/questions/25704853/ unmarshalling-nested-list-of-xml-items-using-jaxb帮助... – deepakl