【发布时间】:2012-12-05 11:17:56
【问题描述】:
大家好,我使用 restFul web 服务公开了一个服务 服务器端代码是
@RequestMapping(value = "/getPerson", method = RequestMethod.POST)
public ModelAndView getPerson(@RequestParam("inputXml") String inputXml) {
-------------------------
----------------------------------------
}
return new ModelAndView("userXmlView", BindingResult.MODEL_KEY_PREFIX
+ String.class, "测试");
}
客户端实现是:
URL oracle = new URL("http://localhost:8081/testWeb/restServices/getPerson?inputXml=input");
System.out.println("Oracle URl is "+oracle);
HttpURLConnection connection = (HttpURLConnection)oracle.openConnection();
connection.setDoOutput(true);
connection.setRequestProperty("Content-type", "application/xml; charset:ISO-8859-1");
connection.setRequestMethod("POST");
BufferedReader in = new BufferedReader(new InputStreamReader(
connection.getInputStream()));
String inputLine;
while ((inputLine = in.readLine()) != null)
System.out.println(inputLine);
in.close();
我能够使用 URL 访问服务
http://localhost:8081/testWeb/restServices/getPerson?inputXml="input"
实际上我的要求是,我需要像这样将 xml 字符串作为输入传递
http://localhost:8081/testWeb/restServices/getPerson?inputXml="<?xml%20version="1.0"%20encoding="UTF-8"%20standalone="yes"?><product><code>WI1</code><name>Widget%20Number%20One</name><price>300.0</price></product>"
请帮我找到解决办法
【问题讨论】:
标签: web-services spring rest