Question

可能重复：
How to reading XML from a URL in Android

我们的想法是创建一个Android应用程序，它可以与现有服务器一起发送XML消息并获取XML响应。

以下是代码：

try {
                URL url1 = null;
                url1 = new URL("http://mysite.com/");
                URLConnection urlc1 = url1.openConnection();
                HttpURLConnection con1 = (HttpURLConnection) urlc1;

                con1.setRequestMethod("POST");

                con1.setUseCaches(false);
                con1.setDoOutput(true);
                con1.setDoInput(true);
                con1.setRequestProperty("accept-charset", "UTF-8");
                con1.setRequestProperty("content-type", "application/x-www-form-urlencoded");


                XmlSerializer ser1 = Xml.newSerializer();
                StringWriter writer = new StringWriter();
                ser1.setOutput(writer);
                ser1.startDocument("UTF-8", true);              
                ser1.startTag("", "request");
                ser1.attribute("", "query", "open_session");
                ser1.startTag("", "");
                ser1.attribute("", "client", "name");
                ser1.attribute("", "version", "1.0");
                ser1.endTag("", "");
                ser1.endTag("", "request");
                ser1.endDocument();

                String query = writer.toString();   

                con1.connect();

                OutputStreamWriter out1 = new OutputStreamWriter(con1.getOutputStream());  
                out1.write(query);  
                out1.flush();   


                String result1 = null;
                BufferedReader br1 = new BufferedReader(new InputStreamReader(con1.getInputStream()));
                 StringBuffer sb = new StringBuffer();
                 String line;
                 while ((line = br1.readLine()) != null){
                     sb.append(line);
                     }
                 br1.close();
                 result1 = sb.toString();


            } catch (MalformedURLException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

程序打开连接，创建登录请求XML，将其发送到服务器并获得XML答案。

但它无法正常工作。我得到的回应是：

<html><head><title>302 Found</title></head></body bg color="white"><center><h1>302 Found</h1></center><hr><center>nginx/1.0.0</center></body><html>

这不是服务器列出的错误。成功的答案应该是：

<result success="true">
<session-info time-out="number" id="number" start="number"/>
</result>

有趣的是，当我评论这部分代码时：

OutputStreamWriter out1 = new OutputStreamWriter(con1.getOutputStream());  
out1.write(query);  
out1.flush();

我得到了正确的XML错误答案：

<result success="false">
<error error-code="XML-01">
Wrong XML
</error>
</result>

你能帮帮我吗？

提前致谢。

Answer 1

302 Found是重定向消息，如果您使用HttpClient，则可以在内部处理这些重定向。

查看https://stackoverflow.com/a/2560129/1321873，了解如何将XML发布到服务器。（虽然他们发送的是SOAP XML，但发送普通XML的机制也是一样的 - 只需更改内容类型）

Android客户端使用XML消息

1 个答案: