如何格式化restful web服务的xml输入并调用服务

Question

嗨，我使用restFul web服务服务器端代码公开的服务是

@requestmapping(value=“/getPerson”,method=requestmethod.post）public ModelAndView getPerson(@requestparam（“inputXml”）String inputXml）{
----------------------------------
}返回新的ModelAndView（“userxmlview”,bindingresult.model_key_prefix+String.class,“test”）；}

客户端实现为：

        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(); 

我可以使用URLhttp://localhost:8081/testweb/restservices/getperson访问服务？inputxml=“input”实际上，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

Answer 1

Maya,/getPerson不是RESTful URI名称。您应该使用类似/person的代码。这样，您就可以使用HTTP获取或删除。