Android，通过 HTTP POST (SOAP) 发送 XML
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【中文标题】Android，通过 HTTP POST (SOAP) 发送 XML【英文标题】：Android, sending XML via HTTP POST (SOAP) 【发布时间】：2011-02-03 08:02:23 【问题描述】：我想通过 android 调用网络服务。我需要通过 HTTP 将一些 XML 发布到 URL。 我发现这是为了发送 POST 而剪掉的，但我不知道如何包含/添加 XML 数据本身。
public void postData()
// Create a new HttpClient and Post Header
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://10.10.4.35:53011/");
try
// Add your data
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("Content-Type", "application/soap+xml"));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
// Where/how to add the XML data?
// Execute HTTP Post Request
HttpResponse response = httpclient.execute(httppost);
catch (ClientProtocolException e)
// TODO Auto-generated catch block
catch (IOException e)
// TODO Auto-generated catch block
这是我需要模仿的完整 POST 消息：
POST /a8103e90-f1e3-11dd-bfdb-8b1fcff1a110 HTTP/1.1
Host: 10.10.4.35:53011
Content-Type: application/soap+xml
Content-Length: 602
<?xml version='1.0' encoding='UTF-8' ?>
<s12:Envelope xmlns:s12="http://www.w3.org/2003/05/soap-envelope" xmlns:wsa="http://schemas.xmlsoap.org/ws/2004/08/addressing">
<s12:Header>
<wsa:MessageID>urn:uuid:fc061d40-3d63-11df-bfba-62764ccc0e48</wsa:MessageID>
<wsa:Action>http://schemas.xmlsoap.org/ws/2004/09/transfer/Get</wsa:Action>
<wsa:To>urn:uuid:a8103e90-f1e3-11dd-bfdb-8b1fcff1a110</wsa:To>
<wsa:ReplyTo>
<wsa:Address>http://schemas.xmlsoap.org/ws/2004/08/addressing/role/anonymous</wsa:Address>
</wsa:ReplyTo>
</s12:Header>
<s12:Body />
</s12:Envelope>
【问题讨论】：
嗨。你是怎么做到的？我应该放 SOAPRequestXML = "POST /a8103e.... " 还是 ""? @Intosia 是新手，也面临同样的问题。您能否详细解释一下以下解决方案，以便我理解。谢谢 【参考方案1】：-
首先，您可以为此 SOAP 请求创建一个字符串模板，并在此模板中替换用户在运行时提供的值以创建一个有效的请求。
将此字符串包装在 StringEntity 中并将其内容类型设置为 text/xml
在 SOAP 请求中设置此实体。
类似：
HttpPost httppost = new HttpPost(SERVICE_EPR);
StringEntity se = new StringEntity(SOAPRequestXML,HTTP.UTF_8);
se.setContentType("text/xml");
httppost.setHeader("Content-Type","application/soap+xml;charset=UTF-8");
httppost.setEntity(se);
HttpClient httpclient = new DefaultHttpClient();
BasicHttpResponse httpResponse =
(BasicHttpResponse) httpclient.execute(httppost);
response.put("HTTPStatus",httpResponse.getStatusLine().toString());
【讨论】：
你也应该考虑阅读这个：***.com/questions/297586/… 感谢工作，更干净，虽然我不得不添加一个 'HttpClient httpclient = new DefaultHttpClient();' 当然，我只是粘贴了与我提出的观点相关的部分……很高兴知道它有帮助。干杯！ 嗨。我应该放 SOAPRequestXML = "POST /a8103e.... " 还是 ""? @Samuh 你能否澄清一下最后一行代码response.put(...); 的作用以及对象response 的类型是什么？【参考方案2】：
这里是发送soap msg的替代方法。
public String setSoapMsg(String targetURL, String urlParameters)
URL url;
HttpURLConnection connection = null;
try
//Create connection
url = new URL(targetURL);
// for not trusted site (https)
// _FakeX509TrustManager.allowAllSSL();
// System.setProperty("javax.net.debug","all");
connection = (HttpURLConnection)url.openConnection();
connection.setRequestMethod("POST");
connection.setRequestProperty("SOAPAction", "**** SOAP ACTION VALUE HERE ****");
connection.setUseCaches (false);
connection.setDoInput(true);
connection.setDoOutput(true);
//Send request
DataOutputStream wr = new DataOutputStream (
connection.getOutputStream ());
wr.writeBytes (urlParameters);
wr.flush ();
wr.close ();
//Get Response
InputStream is ;
Log.i("response", "code="+connection.getResponseCode());
if(connection.getResponseCode()<=400)
is=connection.getInputStream();
else
/* error from server */
is = connection.getErrorStream();
// is= connection.getInputStream();
BufferedReader rd = new BufferedReader(new InputStreamReader(is));
String line;
StringBuffer response = new StringBuffer();
while((line = rd.readLine()) != null)
response.append(line);
response.append('\r');
rd.close();
Log.i("response", ""+response.toString());
return response.toString();
catch (Exception e)
Log.e("error https", "", e);
return null;
finally
if(connection != null)
connection.disconnect();
希望对您有所帮助。如果有人想知道allowAllSSL() 的方法，谷歌它:)。
【讨论】：
你的方法对我有用，虽然我必须添加 'connection.setRequestProperty("Content-Type", "text/xml");'因为url参数是soap请求中的xml【参考方案3】：所以如果你使用：
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
它仍然是休息，但如果你使用：
StringEntity se = new StringEntity(SOAPRequestXML,HTTP.UTF_8);
httppost.setEntity(se);
这是肥皂？？？
【讨论】：
【参考方案4】：通过 http POST 将 XML 发送到 WS 的示例。
DefaultHttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://foo/service1.asmx/GetUID");
//XML example to send via Web Service.
StringBuilder sb = new StringBuilder();
sb.append("<myXML><Parametro><name>IdApp</name><value>1234567890</value></Parameter>");
sb.append("<Parameter><name>UID1</name><value>abc12421</value></Parameter>");
sb.append("</myXML>");
httppost.addHeader("Accept", "text/xml");
httppost.addHeader("Content-Type", "application/x-www-form-urlencoded");
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1);
nameValuePairs.add(new BasicNameValuePair("myxml", sb.toString());//WS Parameter and Value
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
【讨论】：
【参考方案5】：这是我发送 html 的代码.... 可以看到数据是nameValuePairs.add(...)
HttpClient httpclient = new DefaultHttpClient();
// Your URL
HttpPost httppost = new HttpPost("http://192.71.100.21:8000");
try
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
// Your DATA
nameValuePairs.add(new BasicNameValuePair("id", "12345"));
nameValuePairs.add(new BasicNameValuePair("stringdata","AndDev is Cool!"));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response;
response = httpclient.execute(httppost);
catch (ClientProtocolException e)
// TODO Auto-generated catch block
e.printStackTrace();
catch (IOException e)
// TODO Auto-generated catch block
e.printStackTrace();
【讨论】：
【参考方案6】：我也必须在 Android 上通过 HTTP Post 发送一些 XML。
String xml = "xml-block";
StringEntity se = new StringEntity(xml,"UTF-8");
se.setContentType("application/atom+xml");
HttpPost postRequest = new HttpPost("http://some.url");
postRequest.setEntity(se);
希望它有效！
【讨论】：
【参考方案7】：这里是代码的 sn-ps 代码，我用于在 SOAP 服务中发布 xml 并作为回报从 web 获取 Inputstream。
private InputStream call(String soapAction, String xml) throws IOException
byte[] requestData = xml.getBytes("UTF-8");
URL url = new URL(URL);
connection = (HttpURLConnection) url.openConnection();
connection.setRequestProperty("Accept-Charset", "UTF-8");
// connection.setRequestProperty("Accept-Encoding","gzip,deflate");
connection.setRequestProperty("Content-Type", "text/xml; UTF-8");
connection.setRequestProperty("SOAPAction", soapAction);
connection.setRequestProperty("User-Agent", "android");
connection.setRequestProperty("Host",
"base_urlforwebservices like - xyz.net");
// connection
// .setRequestProperty("Content-Length", "" + requestData.length);
connection.setRequestMethod("POST");
connection.setDoOutput(true);
connection.setDoInput(true);
os = connection.getOutputStream();
os.write(requestData, 0, requestData.length);
os.flush();
os.close();
is = connection.getInputStream();
return is; // inputStream
这里的xml：就是构建的xml请求，用来调用服务。
玩得开心；
【讨论】：
抱歉，我找不到 URL 类。它在哪里？它是标准 Android Java 类的一部分吗？ 它的 here 是的，它在 android 中，因为 api 级别 1。 你的问题解决了吗？？ 嗯，一年，我采取了另一种方法，但关于找不到这门课，我只是瞎了眼！谢谢！ :-)【参考方案8】：另一种方法是使用Apache Call。需要提供Api URL、Action URI和API Body
InputStream input = new ByteArrayInputStream(apiBody.getBytes());
Service service = new Service();
Call call = (Call) service.createCall();
SOAPEnvelope soapEnvelope = new SOAPEnvelope(input);
call.setTargetEndpointAddress(new URL(apiUrl));
call.setUseSOAPAction(true);
if(StringUtils.isNotEmpty(actionURI))
call.setSOAPActionURI(actionURI);
soapEnvelope = call.invoke(soapEnvelope);
return soapEnvelope.toString();
【讨论】：
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