我正在使用以下代码从服务器获取数据，如果来自服务器的值为GEEK，那么它将加载到下一个类，但未加载新视图。能说说是什么问题吗？
【问题讨论】：
【参考方案1】：参考此代码........，使用修剪功能，因为它消除了字符串中的空间..
public class HttpLogin extends Activity
/** Called when the activity is first created. */
private Button login;
private EditText username, password;
@Override
public void onCreate(Bundle savedInstanceState)
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
login = (Button) findViewById(R.id.login);
username = (EditText) findViewById(R.id.username);
password = (EditText) findViewById(R.id.password);
login.setOnClickListener(new OnClickListener()
@Override
public void onClick(View v)
String Re;
String mUsername = username.getText().toString();
String mPassword = password.getText().toString();
Re=tryLogin(mUsername, mPassword);
Log.d(" Check ","Here");
Log.d("Re",Re);
String temp_check=Re.trim();
if(temp_check.equals("GEEK"))
Intent i = new Intent();
i.setClassName(v.getContext(),"com.httplogin.MainScreen");
startActivity(i);
else
//Toast.makeText(HttpLogin.this,"MAX Returned",0).show();
displayAlert(Re);
// Intent i = new Intent();
// i.setClassName(v.getContext(),"com.san.MainScreen");
// startActivity(i);
);
protected String tryLogin(String mUsername, String mPassword)
Log.d(" TryLoginCheck ","Here");
HttpURLConnection connection;
OutputStreamWriter request = null;
URL url = null;
String response = null;
String temp=null;
String parameters = "username="+mUsername+"&password="+mPassword;
System.out.println("UserName"+mUsername+"\n"+"password"+mPassword);
Log.d("Parameters",parameters);
try
url = new URL("http://serverspace/script.php");
connection = (HttpURLConnection) url.openConnection();
connection.setDoOutput(true);
connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
connection.setRequestMethod("POST");
request = new OutputStreamWriter(connection.getOutputStream());
request.write(parameters);
request.flush();
request.close();
String line = "";
InputStreamReader isr = new InputStreamReader(connection.getInputStream());
BufferedReader reader = new BufferedReader(isr);
StringBuilder sb = new StringBuilder();
while ((line = reader.readLine()) != null)
sb.append(line + "\n");
temp=sb.toString();
Log.d("Temp",temp);
// Response from server after login process will be stored in response variable.
response = sb.toString();
Log.d("Response",response);
Log.d("Sb Value",sb.toString());
isr.close();
reader.close();
catch(IOException e)
Toast.makeText(this,e.toString(),0).show();
// Log.d("Response",response);
return response;
public void displayAlert(String Re)
new AlertDialog.Builder(this).setMessage(Re)
.setTitle("Returned Value")
.setCancelable(true)
.setNeutralButton(android.R.string.ok, new DialogInterface.OnClickListener()
public void onClick(DialogInterface dialog, int whichButton)
finish();
)
.show();
【讨论】：
【参考方案2】：区分大小写吗？在比较大小写无关紧要的字符串时，您应该使用 .equalsIgnoreCase() ，而在比较大小写时应该使用 .equals() 。您是否也已逐步完成并调试以确保 tryLogin(mUsername, mPassword) 返回预期值？
【讨论】：
if(Re.equals("GEEK"))... 在这里阅读更多-> developer.android.com/reference/java/lang/String.html【参考方案3】：关于您的代码需要注意的重要一点是，您永远不应该从 UI 线程运行长时间的操作，例如远程服务器访问（在您的情况下是 tryLogin() 调用）。这种编程会导致您的应用程序出现 ANR。有关此主题的更多详细信息，请阅读this article
简单地说，不是从 onCreate 调用 tryLogin，而是创建异步任务
new LoginTask().execute(mUsername, mPassword);
登录任务应该这样定义：
private class LoginTask extends AsyncTask<string, void,="" String="">
protected String doInBackground(String... login)
return tryLogin((login[0],login[1]);
protected void onPostExecute(String result)
if(result.equalsIgnoreCase("GEEK"))
Intent i = new Intent();
i.setClassName(v.getContext(),"com.httplogin.MainScreen");
startActivity(i);
【讨论】：
见我上面在我的回答中添加的例子。