Question

我正在使用android studio创建一个Android应用程序。登录页面使用PHP检查存储在MYSQL数据库中的详细信息的用户用户名和密码详细信息，并向用户提供有关输入的详细信息是成功还是不成功的信息。如果登录成功，我目前无法打开新活动。有人可以帮我这个吗？谢谢。

public class BackgroundWorker extends AsyncTask<String,Void,String> {
    Context context;
    AlertDialog alertDialog;
    BackgroundWorker (Context ctx) {
        context = ctx;
    }
    @Override
    protected String doInBackground(String... params) {
        String type = params[0];
        String login_url = "http://192.168.1.10/stafflogin.php";
        String createAccount_url = "http://192.168.1.10/staffcreateaccount.php";
        if(type.equals("login")) {
            try {
                String user_name = params[1];
                String password = params[2];
                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 post_data = URLEncoder.encode("user_name","UTF-8")+"="+URLEncoder.encode(user_name,"UTF-8")+"&"
                        +URLEncoder.encode("password","UTF-8")+"="+URLEncoder.encode(password,"UTF-8");
                bufferedWriter.write(post_data);
                bufferedWriter.flush();
                bufferedWriter.close();
                outputStream.close();
                InputStream inputStream = httpURLConnection.getInputStream();
                BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream,"iso-8859-1"));
                String result="";
                String line="";
                while((line = bufferedReader.readLine())!= null) {
                    result += line;
                }
                bufferedReader.close();
                inputStream.close();
                httpURLConnection.disconnect();
                return result;
            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return null;
    }

    @Override
    protected void onPreExecute() {
        alertDialog = new AlertDialog.Builder(context).create();
        alertDialog.setTitle("Login Status");
    }

    @Override
    protected void onPostExecute(String result) {
        alertDialog.setMessage(result);
        alertDialog.show();


    }

    @Override
    protected void onProgressUpdate(Void... values) {
        super.onProgressUpdate(values);
    }

}

Answer 1

在onPostExecute方法中添加此

if(result!=null && result.equals("successful")){
Intent intent = new Intent(LoginActivity.this,NextActivity.class);
startActivity(intent);
}

我编写了示例代码。请在“成功”，LoginActivity，NextActivity中使用正确的值。

Answer 2

正如您所说，您正在收到消息...成功登录或未成功然后你可以打开像...这样的新活动在onPostExecute（）..

if(result!=null && result.equalsIgnoreCase(message from PHP on succesful))
{
 Intent intent = new Intent(context, NextPage.class);
            startActivity(intent);
            finish();
}

希望这会对你有帮助..

Answer 3

假设您将获得成功响应代码200（状态正常）。这是你可以尝试的。

    public class BackgroundWorker extends AsyncTask<String, Void, String> {
    ...
    BackgroundWorker (Context ctx) {
    context = ctx.getApplicationContext();
    }
    boolean statusOk = false;

        @Override
        protected String doInBackground(String... params) {
            ...
            InputStream inputStream = httpURLConnection.getInputStream();
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream, "iso-8859-1"));
            String result = "";
            String line = "";
            int statusCode = httpURLConnection.getResponseCode(); //get response code

            if (statusCode == HttpURLConnection.HTTP_OK) {
                while ((line = bufferedReader.readLine()) != null) {
                    result += line;
                }
                bufferedReader.close();
                statusOk = true;
                return result;
            }
        }

        @Override
        protected void onPostExecute(String result) {
            if (statusOk) {

                Intent intent = new Intent(context, MainPage.class);// replace  MainPage with the activity you want to start
                context.startActivity(intent);
            }

        }
    }

Answer 4

由于您在运行php文件后获得了结果，因此您可以使用onPostExecute()方法启动新的activity。如果您愿意，可以使用以下data

将activity传递给将在此处打开的新code

 @Override
    protected void onPostExecute(String result) {
       if(result!=null && results.equals("what ever you receive from the php"){
           Intent i= new Intent(this,theNameOfTheActivityToBeOpen.class);
           i.putExtra("nameYouWantToReferInOtherActivity",result);
           startActivity(i);
    }
   else{
         }
            alertDialog.setMessage(result);
            alertDialog.show();

并接收发送到新活动的内容使用以下代码

 Bundle anyNameForYourBundle=getIntent().getExtras();
  //since a String is passed
   String temp = anyNameForYourBundle.getString("nameYouWantToReferInOtherActivity");

Answer 5

试试这是工作

 public class BackgroundWorker extends AsyncTask<String, Void, String> {
  ...
BackgroundWorker (Context ctx) {
context = ctx.getApplicationContext();
}
boolean statusOk = false;

    @Override
    protected String doInBackground(String... params) {
        ...
        InputStream inputStream = httpURLConnection.getInputStream();
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream, "iso-8859-1"));
        String result = "";
        String line = "";
        int statusCode = httpURLConnection.getResponseCode(); //get response code

        if (statusCode == HttpURLConnection.HTTP_OK) {
            while ((line = bufferedReader.readLine()) != null) {
                result += line;
            }
            bufferedReader.close();
            statusOk = true;
            return result;
        }
    }

    @Override
    protected void onPostExecute(String result) {
        if (statusOk) {

            Intent intent = new Intent(context, MainPage.class);// replace  MainPage with the activity you want to start
            context.startActivity(intent);
        }

    }
}

成功登录android studio app后打开一个新活动

5 个答案: