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我想通过http获取音频文件从安全的宁静服务获取,我已经成功地接收和解析文本XML服务,但有点困惑,如何处理音频文件。如何通过HTTP获取音频文件?
代码调用安全RESTful服务和XML响应
String callWebService(String serviceURL) {
// http get client
HttpClient client = getClient();
HttpGet getRequest = new HttpGet();
try {
// construct a URI object
getRequest.setURI(new URI(serviceURL));
} catch (URISyntaxException e) {
Log.e("URISyntaxException", e.toString());
}
// buffer reader to read the response
BufferedReader in = null;
// the service response
HttpResponse response = null;
try {
// execute the request
response = client.execute(getRequest);
} catch (ClientProtocolException e) {
Log.e("ClientProtocolException", e.toString());
} catch (IOException e) {
Log.e("IO exception", e.toString());
}
try {
in = new BufferedReader(new InputStreamReader(response.getEntity()
.getContent()));
} catch (IllegalStateException e) {
Log.e("IllegalStateException", e.toString());
} catch (IOException e) {
Log.e("IO exception", e.toString());
}
StringBuffer buff = new StringBuffer("");
String line = "";
try {
while ((line = in.readLine()) != null) {
buff.append(line);
}
} catch (IOException e) {
Log.e("IO exception", e.toString());
return e.getMessage();
}
try {
in.close();
} catch (IOException e) {
Log.e("IO exception", e.toString());
}
// response, need to be parsed
return buff.toString();
}