我正在尝试向POST
应用中的服务器发出Android
请求,但它没有发生。以下是代码 -
try{
View p = (View) v.getRootView();
EditText usernamefield = (EditText)p.findViewById(R.id.username);
String username = usernamefield.getText().toString();
EditText passwordfield = (EditText)p.findViewById(R.id.pass);
String password = passwordfield.getText().toString();
String apiKey = "ac96d760cb3c33a1ee988750b0b2fd12";
String secret = "cd9118e8d1d32d003e0ed54a202c2bf8";
Log.i(TAG,password);
String authToken = computeMD5hash(username.toLowerCase()).toString()+computeMD5hash(password).toString();
String authSig = computeMD5hash("api_key"+apiKey+"authToken"+authToken+"method"+"auth.getMobileSession"+"username"+username+secret).toString();
Log.i(TAG,authToken);
HttpClient client = new DefaultHttpClient();
Log.i(TAG,"after client1");
HttpPost post = new HttpPost("http://ws.audioscrobbler.com/2.0/");
Log.i(TAG,"after client2");
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("method", "auth.getMobileSession"));
nameValuePairs.add(new BasicNameValuePair("api_key", apiKey));
nameValuePairs.add(new BasicNameValuePair("api_sig", authSig));
nameValuePairs.add(new BasicNameValuePair("format", "json"));
nameValuePairs.add(new BasicNameValuePair("authToken", authToken));
nameValuePairs.add(new BasicNameValuePair("username", username));
post.setEntity(new UrlEncodedFormEntity(nameValuePairs));
Log.i(TAG,post.getURI().toString()); //logs the URL
HttpResponse response = client.execute(post);
int status = response.getStatusLine().getStatusCode();
Log.i(TAG,"Status code is"+status);
Log.i(TAG,"after post");
InputStream ips = response.getEntity().getContent();
BufferedReader buf = new BufferedReader(new InputStreamReader(ips,"UTF-8"));
if(response.getStatusLine().getStatusCode()!= org.apache.commons.httpclient.HttpStatus.SC_OK)
{
Log.i(TAG,"bad http response");
Toast.makeText(getApplicationContext(),"bad httpcode",Toast.LENGTH_LONG).show();
throw new Exception(response.getStatusLine().getReasonPhrase());
}
StringBuilder sb = new StringBuilder();
String s;
while(true)
{
s = buf.readLine();
if(s==null || s.length()==0)
break;
sb.append(s);
}
buf.close();
ips.close();
System.out.print(sb.toString());
}
catch (ClientProtocolException e) {
// TODO Auto-generated catch block
} catch (IOException e) {
// TODO Auto-generated catch block
}
catch(NoSuchAlgorithmException e)
{
}
catch(Exception e){
}
代码执行到Log.i(TAG,post.getURI().toString())
日志语句。它会打印出来的网址 - http://ws.audioscrobbler.com/2.0/。没有附加参数(这很奇怪)。
我不知道使用NameValuePairs将参数添加到URL的实现有什么问题。
答案 0 :(得分:1)
我有一个简单的方法将数据发布到服务器。请使用它并告诉我这是否对您有用:
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("method", "auth.getMobileSession"));
nameValuePairs.add(new BasicNameValuePair("api_key", apiKey));
nameValuePairs.add(new BasicNameValuePair("api_sig", authSig));
nameValuePairs.add(new BasicNameValuePair("format", "json"));
nameValuePairs.add(new BasicNameValuePair("authToken", authToken));
nameValuePairs.add(new BasicNameValuePair("username", username));
//call to method
JSONObject obj = makeHttpRequest(nameValuePairs, "http://ws.audioscrobbler.com/2.0/", "POST");
public static JSONObject makeHttpRequest(List<NameValuePair> params, String url, String method) {
InputStream is = null;
JSONObject jObj = null;
String json = "";
// Making HTTP request
try {
// check for request method
if(method == "POST"){
// request method is POST
// defaultHttpClient
url = url.trim();
Log.e("FETCHING_DATA_FROM",""+url.toString());
HttpPost httpPost = new HttpPost(url);
HttpParams httpParameters = new BasicHttpParams();
// Set the timeout in milliseconds until a connection is established.
int timeoutConnection = 600000;
HttpConnectionParams.setConnectionTimeout(httpParameters, timeoutConnection);
// Set the default socket timeout (SO_TIMEOUT)
// in milliseconds which is the timeout for waiting for data.
int timeoutSocket = 600000;
HttpConnectionParams.setSoTimeout(httpParameters, timeoutSocket);
DefaultHttpClient httpClient = new DefaultHttpClient(httpParameters);
httpPost.setEntity(new UrlEncodedFormEntity(params,"utf-8"));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}else if(method == "GET"){
// request method is GET
if(params!=null){
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
}
HttpGet httpGet = new HttpGet(url);
Log.e("FETCHING_DATA_FROM",""+url.toString());
HttpParams httpParameters = new BasicHttpParams();
// Set the timeout in milliseconds until a connection is established.
int timeoutConnection = 600000;
HttpConnectionParams.setConnectionTimeout(httpParameters, timeoutConnection);
// Set the default socket timeout (SO_TIMEOUT)
// in milliseconds which is the timeout for waiting for data.
int timeoutSocket = 600000;
HttpConnectionParams.setSoTimeout(httpParameters, timeoutSocket);
DefaultHttpClient httpClient = new DefaultHttpClient(httpParameters);
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
return jObj;
}
答案 1 :(得分:0)
您必须将此代码包装在try catch块中,因为此处抛出了几个可能的异常。至于哪个是我不确定的问题,但是这里有一些可能导致问题的常见问题,你需要提供更多信息才能看到它是哪一个: 1)在较新版本的Android上,如果你在主UI线程中进行thisw调用,它将抛出一个例外NetworkOnMainThread异常。您必须在后台线程上执行网络代码。 2)您没有在清单中声明Internet权限,因此会引发安全性异常。
您需要查看logcat中的try / catch的catch部分中的异常或中断。如果您的捕获物如下:
catch(Exception e)
{
}
然后它会默默地吃掉异常并且没有给出任何问题的迹象。
答案 2 :(得分:0)
试试JsonStringer.Like:
HttpPost request = new HttpPost("http://ws.audioscrobbler.com/2.0/something_here");
request.setHeader("Accept", "application/json");
request.setHeader("Content-type", "application/json");
JSONStringer vm;
try {
vm = new JSONStringer().object().key("method")
.value("auth.getMobileSession").key("api_key").value(apikey)
.key("api_sig") .value(authSig).key("format").value("json").key(authToken).value(authToken).key("username").value(username)
.endObject();
StringEntity entity = new StringEntity(vm.toString());
request.setEntity(entity);
HttpClient httpClient = new DefaultHttpClient();
HttpResponse response = httpClient.execute(request);
同样,Kaediil说使用你的回复代码制作try和catch子句。