我使用AndroidStudio开发AndroidApps我开始做一个简单的HttpPost请求我遇到了问题,我发现的所有帖子都是这样做的:
private void CheckLoguin_Request(String User, String Pass){
//Declaration of variables
HttpClient httpClient = new DefaultHttpClient();
HttpPost Request = new HttpPost(url_Loguin);
HttpResponse Response;
List<NameValuePair> BodyRequest_Elements = new ArrayList<NameValuePair>();
BodyRequest_Elements.add(new BasicNameValuePair("user_name", User));
BodyRequest_Elements.add(new BasicNameValuePair("user_passwd", Pass));
Request.setEntity(new UrlEncodedFormEntity(BodyRequest_Elements));
Response = httpClient.execute(Request);
// writing response to log
Log.d("Http Response:", Response.toString());
}
但是当我尝试调试App Android Studio时,在这行中给出了2个错误:
new UrlEncodedFormEntity(BodyRequest_Elements) //Error:(40, 27) error: unreported exception UnsupportedEncodingException; must be caught or declared to be thrown
Response = httpClient.execute(Request); //Error:(41, 38) error: unreported exception IOException; must be caught or declared to be thrown
我可能需要安装更多库或支持库吗?我做得不好?有人可以帮帮我吗?提前致谢,对不起我的英语!
PD1:如果您需要更多信息或代码,请告诉我!
答案 0 :(得分:1)
2016年更新
使用 HttpURLConnection 类,我们需要打开 BufferedWritter 以插入我们的实体值,如下代码所示:
//Declaration of variables
private List<NameValuePair> ListOfValues;
...
public Check_Loguin_Request(Context cx,String url, List<NameValuePair> ListOfValues)
{
this.cx = cx;
this.Url = url;
this.ListOfValues= ListOfValues;
}
@Override
protected String doInBackground(String... strings) {
//Declaration of variables
InputStream is = null;
String Result = "";
URL urll = new URL(url);
HttpURLConnection conn = (HttpURLConnection) urll.openConnection();
conn.setReadTimeout(10000);
conn.setConnectTimeout(15000);
conn.setRequestMethod("POST");
conn.setDoInput(true);
conn.setDoOutput(true);
OutputStream os = conn.getOutputStream();
BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(os, "UTF-8"));
writer.write(Build_HttpBody(ListOfValues));
writer.flush();
writer.close();
os.close();
// Starts request
conn.connect();
int ResponseCode = conn.getResponseCode();
if (ResponseCode == 200) {
is = conn.getInputStream();
String EntityResult = ReadResponse_HttpURLConnection(is);
}
else {
throw new RuntimeException("Invalid Status Code");
}
}
注意不推荐使用DefaultHttpClient类(2011)
有关此link之后的详细信息。
尝试使用以下代码,请记住这种情况始终使用AsyncTask:
private class Check_Loguin_Request extends AsyncTask <String,Void,String>{
Context cx;
String Url;
List<NameValuePair> BodyRequest_Elements;
public Check_Loguin_Request(Context cx,String url, List<NameValuePair> ListOfValues)
{
this.cx = cx;
this.Url = url;
this.BodyRequest_Elements = ListOfValues;
}
private String convertStreamToString(InputStream is) {
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();
String line = null;
try {
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
is.close();
} catch (IOException e) {
e.printStackTrace();
}
}
return sb.toString();
}
@Override
protected String doInBackground(String... strings) {
//Declaration of variables
DefaultHttpClient httpClient;
HttpPost Request = new HttpPost(url_Loguin);
HttpResponse Response;
HttpParams httpParameters = new BasicHttpParams();
httpParameters.setParameter(CoreProtocolPNames.PROTOCOL_VERSION, HttpVersion.HTTP_1_1);
// Set the timeout in milliseconds until a connection is established.
// The default value is zero, that means the timeout is not used.
int timeoutConnection = 3000;
HttpConnectionParams.setConnectionTimeout(httpParameters, timeoutConnection);
// Set the default socket timeout (SO_TIMEOUT)
// in milliseconds which is the timeout for waiting for data.
int timeoutSocket = 5000;
HttpConnectionParams.setSoTimeout(httpParameters, timeoutSocket);
httpClient = new DefaultHttpClient(httpParameters);
try {
HttpEntity entity = new UrlEncodedFormEntity(BodyRequest_Elements);
Request.setHeader(entity.getContentType());
Request.setEntity(entity);
Response = httpClient.execute(Request);
if(Response.getStatusLine().getStatusCode() == 200){
String EntityResult = EntityUtils.toString(Response.getEntity());
//HttpEntity EntityResult = Response.getEntity();
//InputStream iStream = EntityResult.getContent();
//JSONObject json = new JSONObject(convertStreamToString(iStream));
EntityResult = EntityResult.replaceAll("[()]", "");
JSONObject json = new JSONObject(EntityResult);
String Result = json.optString("code").toString();
return Result;
}
else{
throw new RuntimeException("Invalid Status Code");
}
}
catch (Exception ex){
Log.getStackTraceString(ex);
return ex.toString();
}
}
}
答案 1 :(得分:0)
您显然没有处理从构造函数抛出的IO异常:
请参阅:Android docs on UrlEncodedFormEntity
让我感到困惑的是,你的IDE没有报告这个,或者这个代码甚至是如何编译的,更不用说运行/调试了。
编辑: 如果你是这个东西的新手,也许值得一看: Android Volley,这是Google为Android网络开发的一个非常简单但有效的库。