Php 如何在android json中成功登录？

我正在使用登录和注册，我创建了登录类并进行了一些编码。当我尝试注册时，它会给我错误，我是android新手，所以请帮助。。 这是我的登录类：

public class Login extends AppCompatActivity {

    private ProgressDialog pDialog;

    Button login;
    private static final String TAG_SUCCESS = "success";
    JSONParser jsonParser = new JSONParser();
    private int serverResponseCode = 0;
    TextView tac1;
EditText email,pass;
    private static String url_create_book = "http://cloud......com/broccoli/login.php";

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_login);
        Toolbar toolbar = (Toolbar) findViewById(R.id.toolbar);
        setSupportActionBar(toolbar);
        getSupportActionBar().setDisplayHomeAsUpEnabled(true);


        email=(EditText)findViewById(R.id.email);
        pass=(EditText)findViewById(R.id.password);

        tac1 = (TextView) findViewById(R.id.cAcc);
        tac1.setOnClickListener(new View.OnClickListener() {

            @Override
            public void onClick(View v) {


              startActivity(new Intent(Login.this, RegistrationForm.class));
        }
    });

    login=(Button)findViewById(R.id.lbtn);
    login.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {




            new CreateNewProduct().execute();


        }
    });

}


class CreateNewProduct extends AsyncTask<String, String, String> {
    private  String eml;
    private  String pswrd;



    /**
     * Before starting background thread Show Progress Dialog
     * */
    @Override
    protected void onPreExecute() {
        super.onPreExecute();

       eml = email.getText().toString();
        pswrd = pass.getText().toString();


        //Toast.makeText(RegistrationForm.this,
        //dobr, Toast.LENGTH_SHORT).show();
    }




    protected String doInBackground(String... args) {


        // Building Parameters
        List<NameValuePair> params = new ArrayList<NameValuePair>();

        params.add(new BasicNameValuePair("email", eml));

        params.add(new BasicNameValuePair("password", pswrd));




        // getting JSON Object
        // Note that create product url accepts POST method
        JSONObject json = jsonParser.makeHttpRequest(url_create_book,
                "POST", params);

        // check log cat fro response
        Log.d("Create Response", json.toString());

        // check for success tag
        try {
            int success = json.getInt(TAG_SUCCESS);

            if (success == 1) {
                // successfully created product
                Intent i = new Intent(getApplicationContext(), TypeMenu.class);
                startActivity(i);

                // closing this screen
                finish();
            } else {
                // failed to create product
            }
        } catch (JSONException e) {
            e.printStackTrace();
        }

        return null;
    }

    /**
     * After completing background task Dismiss the progress dialog
     * **/
    protected void onPostExecute(String file_url) {
        // dismiss the dialog once done
        pDialog.dismiss();
    }

}

---

在浏览器中用参数检查你的url。然后检查它是否给出正确的json值。是的，正确。请查看我的php文件。如果我手动匹配电子邮件和密码，那么它会显示“成功”。我不知道哪里错了？？？我可以告诉你通过截取使用此登录API。。如果您感兴趣，请在浏览器中使用参数检查您的url。然后检查它是否提供了正确的json值。是的，这是正确的。请查看我的php文件。如果我手动匹配电子邮件和密码，那么它将显示“成功”。我不知道它错在哪里？？？我可以告诉您通过截取使用此登录API。。如果你有兴趣

<?php 
include ('config.php');



$email = $_POST['email'];
$password = $_POST['password'];

$sql = mysqli_query($conn,"SELECT * FROM UserInfo WHERE email='".$email."' AND password='".$password."'");



$check = mysqli_fetch_array($sql);

if(isset($check)){
echo 'success';
}else{
echo 'failure';
}

mysqli_close($con);
?>

    E/JSON Parser: Error parsing data org.json.JSONException: Value failure of type java.lang.String cannot be converted to JSONObject
E/AndroidRuntime: FATAL EXCEPTION: AsyncTask #3
                  Process: com.example.zeba.broccoli, PID: 24295
                  java.lang.RuntimeException: An error occured while executing doInBackground()
                      at android.os.AsyncTask$3.done(AsyncTask.java:304)
                      at java.util.concurrent.FutureTask.finishCompletion(FutureTask.java:355)
                      at java.util.concurrent.FutureTask.setException(FutureTask.java:222)
                      at java.util.concurrent.FutureTask.run(FutureTask.java:242)
                      at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:231)
                      at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1112)
                      at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:587)
                      at java.lang.Thread.run(Thread.java:818)
                   Caused by: java.lang.NullPointerException: Attempt to invoke virtual method 'java.lang.String org.json.JSONObject.toString()' on a null object reference
                      at com.example.zeba.broccoli.Login$CreateNewProduct.doInBackground(Login.java:125)
                      at com.example.zeba.broccoli.Login$CreateNewProduct.doInBackground(Login.java:82)
                      at android.os.AsyncTask$2.call(AsyncTask.java:292)
                      at java.util.concurrent.FutureTask.run(FutureTask.java:237)
                      at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:231) 
                      at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1112) 
                      at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:587) 
                      at java.lang.Thread.run(Thread.java:818) 
D/mali_winsys: new_window_surface returns 0x3000,  [540x960]-format:1
I/Timeline: Timeline: Activity_idle id: android.os.BinderProxy@369d02de time:39270446
I/Process: Sending signal. PID: 24295 SIG: 9
Application terminated.