Question

如何在登录活动成功后开始活动..我试过这个，how to start an activity after my login Activity但似乎并没有为我工作。我把我的登录活动放在我的主...然后我的警告消息在PHP脚本.. 这是我的backgroundtask编码：

@Override
protected String doInBackground(String... params) {
    String reg_url = "http://10.0.2.2/webapp/register.php";
    String login_url = "http://10.0.2.2/webapp/login.php";
    String method = params[0];
    if(method.equals("register"))
    {
    String name = params[1];
    String user_name = params[2];
    String user_pass = params [3];

        try {
            URL url = new URL(reg_url);
            HttpURLConnection httpURLConnection =(HttpURLConnection) url.openConnection();
            httpURLConnection.setRequestMethod("POST");
            httpURLConnection.setDoOutput(true);
            OutputStream OS = httpURLConnection.getOutputStream();
            BufferedWriter bufferedWriter= new BufferedWriter(new OutputStreamWriter(OS,"UTF-8"));
            String data = URLEncoder.encode("user","UTF-8") + "=" + URLEncoder.encode(name,"UTF-8")+"&"+
                    URLEncoder.encode("user_name","UTF-8") + "=" + URLEncoder.encode(user_name,"UTF-8")+"&"+
                    URLEncoder.encode("user_pass","UTF-8") + "=" + URLEncoder.encode(user_pass,"UTF-8");

            bufferedWriter.write(data);
            bufferedWriter.flush();
            bufferedWriter.close();
            OS.close();
            InputStream IS = httpURLConnection.getInputStream();
            IS.close();
            return "Registration Success...";
        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    else if (method.equals("login"))
    {
       String login_name = params[1];
        String login_pass = params[2];
        try {
            URL url = new URL(login_url);
            HttpURLConnection httpURLConnection =(HttpURLConnection)url.openConnection();
            httpURLConnection.setRequestMethod("POST");
            httpURLConnection.setDoOutput(true);
            httpURLConnection.setDoInput(true);
            OutputStream outputStream = httpURLConnection.getOutputStream();
            BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(outputStream,"UTF-8"));
      String data = URLEncoder.encode("login_name","UTF-8")+"="+URLEncoder.encode(login_name,"UTF-8")+"&"+
              URLEncoder.encode("login_pass","UTF-8")+"="+URLEncoder.encode(login_pass,"UTF-8");
            bufferedWriter.write(data);
            bufferedWriter.flush();
            bufferedWriter.close();
            outputStream.close();

            InputStream inputStream = httpURLConnection.getInputStream();
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream,"iso-8859-1"));
            String response = "";
            String line = "";
            while ((line = bufferedReader.readLine())!=null)
            {
              response+= line ;
            }
            bufferedReader.close();
            inputStream.close();
            httpURLConnection.disconnect();
            return response;

        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    return null;
}

@Override
protected void onPostExecute(String result) {
    if(result.equals("Registration Success..."))
    {
        Toast.makeText(ctx, result, Toast.LENGTH_LONG).show();
    }
    else if(result.contains("Login Success")) 
    {
        Intent i = new Intent(getApplicationContext,NewActivity.class);
        getApplicationContext.startActivity(i);
        alertDialog.setMessage(result);
        alertDialog.show();
    }


 }
}

请帮助我......这是我一直试图解决这个问题的一周.. tq

Answer 1

在登录过程中，您尝试在登录成功时调试“响应”的值（第66行）

这是调试'响应'的两种方法

将调试工具与BreakPoint一起使用
在返回响应之前，将代码放在下面并观察您的Logcat。

Log.e（“CODE”，“响应：”+响应）;

当您知道响应内容时，请将onPostExecute更改为以下代码，并将{$ CONTENTS YOU DEBUGGED}替换为调试结果。

@Override
protected void onPostExecute(String result) {
    if (result.equals("Registration Success...")) {
        Toast.makeText(ctx, result, Toast.LENGTH_LONG).show();
    } else if(result.contains("Hello") { 
        Intent i = new Intent(ctx, SomethingClass.class);
        ctx.startActivity(i);
    } else {
        alertDialog.setMessage(result);
        alertDialog.show();
    }
}

编辑））

响应模式是Hello Welcome + $ user。所以，当响应包含'Hello'时，它将移动另一个Activity。而已。

Answer 2

你必须返回String＆＃34;登录成功＆＃34;在void doInBackGround上，然后在void onPostExecute上创建新条件;

@Override
    protected void onPostExecute(String result) {
        if(result.equals("Registration Success..."))
        {
            Toast.makeText(ctx, result, Toast.LENGTH_LONG).show();
        } else if (result.equals("Login success")) 
        { 
            startActivity(new Intent(yourCURRENTactivity.this, yourNEXTactivity.class)); 
        }
        else
        {
            alertDialog.setMessage(result);
            alertDialog.show();
        }

     }
    }

Answer 3

在您的代码中，您已经在意图中提供了不正确的应用程序上下文。它应该是当前活动然后是下一个切换活动。请参阅以下代码

@Override
protected void onPostExecute(String result) {
if(result.equals("Registration Success..."))
{
    Toast.makeText(ctx, result, Toast.LENGTH_LONG).show();
}
else if(result.contains("Login Success")) 
{
    Intent i = new Intent(currentactivity.this,NewActivity.class);
    getApplicationContext.startActivity(i);
    alertDialog.setMessage(result);
    alertDialog.show();
}


}

登录成功后如何启动另一个活动

3 个答案: