嗨所有我使用restFul Web服务服务器端代码公开服务
@RequestMapping(value = "/getPerson",method = RequestMethod.POST)public ModelAndView getPerson(@RequestParam("inputXml")String inputXml){
返回new ModelAndView("userXmlView",BindingResult.MODEL_KEY_PREFIX String.class,"Test"); }
客户端实施是:
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 http://localhost:8081/testWeb/restServices/getPerson?inputXml="input"
访问该服务实际上我的要求是,我需要传递xml字符串作为这样的输入
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>"
请帮我找到解决方案
2 回答
Maya,
/getPerson
不是RESTful URI名称 . 你应该使用像/person
这样的东西 . 这样,你可以GET
它或DELETE
使用HTTP .看看RestAssured
或 Spring 天RestTemplate