android XML通过HTTP post发送和接收

Question

首先，我尝试能够发送带有以下参数的 HttpResponse：

http://www.syslang.com/frengly/controller?action=translateREST&src=en&dest=iw&text=good&email=YYY&password=XXX

代码如下所示：

DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://www.syslang.com/frengly/controller");   
List<NameValuePair> pairs = new ArrayList<NameValuePair>();  
pairs.add(new BasicNameValuePair("src", "en")); 
pairs.add(new BasicNameValuePair("dest", "iw"));  
pairs.add(new BasicNameValuePair("text", "good"));                  
pairs.add(new BasicNameValuePair("email", "YYY")); 
pairs.add(new BasicNameValuePair("password", "XXX"));
UrlEncodedFormEntity entity = new UrlEncodedFormEntity(pairs,HTTP.UTF_8);  
httpPost.setEntity(entity);  
HttpResponse response = httpClient.execute(httpPost); 
HttpEntity httpEntity = response.getEntity();

API structure is available at http://www.frengly.com/ (under the API tab) 共有 5 个参数（src、dest、text、email、password）。

到目前为止，我每次尝试拨打电话

HttpResponse 响应 = httpClient.execute(httpPost); 我不断收到 IO 异常 :(

之后我应该得到类似这样的结构：

-<root>
  <text>good</text>
  <translation>טוב</translation>
  <translationFramed>טוב|</translationFramed>
  <missing/>
  <existing>good,</existing>
  <stat>1/1</stat>
</root>

我想我会处理这部分来构建 XML 并根据需要解析它

ps：我查过了 Android, send and receive XML via HTTP POST method 和许多其他链接，这对我没有多大帮助。

如果需要我的应用程序中的任何代码行，请告诉我...

提前致谢。

Answer 1

您不需要发布。 您需要改为 GET。

// Construct your request here.
String requestURL= "http://www.syslang.com/frengly/controller?action=translateREST&src=en&dest=iw&text=good&email=YYY&password=XXX"

DefaultHttpClient httpClient = new DefaultHttpClient();
HttpGet httpGet = new HttpGet(requestURL);
HttpResponse response = httpClient.execute(httpGet);

还有你的 AndroidManifest.xml 中的互联网权限

<uses-permission android:name="android.permission.INTERNET"/>
<application
....
....