2016-10-10 55 views
1

这里是我的PHP代码,当我想要做一个自定义的方法,如果(字符串等于成功),但它不工作。没有,如果这个方法返回一个字符串值的响应。我只想做一些方法,如果响应给成功无法在凌空if语句使用onresponse方法

<?php 
    define('HOST','mysql.hostinger.in'); 
    define('USER','example'); 
    define('PASS','example'); 
    define('DB','example'); 

    $con = mysqli_connect(HOST,USER,PASS,DB); 

    $username = $_POST['PHONENO']; 
    $password = $_POST['PASSWORD']; 

    $sql = "SELECT * FROM Prago WHERE PHONENO='$username' and PASSWORD='$password'"; 

    $res = mysqli_query($con,$sql); 

    $check = mysqli_fetch_array($res); 

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

    mysqli_close($con); 
?> 

我凌空串请求类

StringRequest stringRequest=new StringRequest(Request.Method.POST, REGISTER_URL, new Response.Listener<String>() { 

     @Override 
     public void onResponse(String response) { 
      if(response.equals("success")){ 
       Toast.makeText(getApplicationContext(),"DONE",Toast.LENGTH_SHORT).show(); 
      } 
      else 
      { 
       Toast.makeText(getApplicationContext(),"check username and password",Toast.LENGTH_SHORT).show(); 
      } 
      callback.onSuccess(response); 

     } 
    }, new Response.ErrorListener() { 
     @Override 
     public void onErrorResponse(VolleyError error) { 
      Toast.makeText(getApplication(),"error in volley",Toast.LENGTH_SHORT).show(); 
     } 
    }){ 
     @Override 
     protected Map<String, String> getParams() throws AuthFailureError { 
      Map<String,String> map=new HashMap<>(); 
      map.put("PHONENO",Number); 
      map.put("PASSWORD",Pass); 
      return map; 
     } 
    }; 

    VolleySingleton.getInstance().addToRequestQueue(stringRequest); 
} 
+0

尝试把一个断点,看看你得到respose。 –

+0

http://blog.strv.com/debugging-in-android-studio-as/ –

+0

@RissmonSuresh先生我没有得到你?你可以解释吗? –

回答

0

由于@greenapps在评论改变.equals().contains()工作,因为在我的情况下的响应指出与双来临引用,所以“成功”!=成功,但“成功”包含成功

0

首先检查通过调试或通过简单地烘烤您的回复收到的回复