JSONException：值

Question

I know that this kind of problem has some solutions and applied some solutions for this problem but I can't solve and i'm confused. Please help me. Here is code:

protected String doInBackground(Boolean... params) {

        String result = null;

        StringBuilder sb = new StringBuilder();


        try {

            // http post
            HttpClient httpclient = new DefaultHttpClient();



            HttpGet httppost = new HttpGet(
                    "http://192.168.2.245/getProducts.php?login=1&user_name=UserName&password=Password");  

            HttpResponse response = httpclient.execute(httppost);
            if (response.getStatusLine().getStatusCode() != 200) {
                Log.d("MyApp", "Server encountered an error");
            }



            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    response.getEntity().getContent(), "utf-8"), 8); //old charset iso-8859-1

            sb = new StringBuilder();

            sb.append(reader.readLine() + "\n");

            String line = null;

            while ((line = reader.readLine()) != null) {

                sb.append(line + "\n");


            }
            result = sb.toString();

            Log.d("test", result);

        } 
        catch (Exception e) {

        Log.e("log_tag", "Error converting result " + e.toString());

        }

        return result;
    }

PHP CODE:

$login=$_GET["login"]; 
$user_name=$_GET["user_name"]; 
$password=$_GET["password"]; 
$output=array(); 
if ($login) { 
$sql=mysql_query("SELECT user_id FROM users WHERE user_name='".$user_name."' AND user_pass='".$password."' "); 

while($row=mysql_fetch_array($sql)) { 
 $user_id=$row["user_id"]; 
} 

$sql=mysql_query("SELECT name,device_id,lat,lon FROM devices WHERE user_id='".$user_id."' LIMIT 100"); 

while($row=mysql_fetch_assoc($sql)) { 
$output[]=$row; } 
} 

print(json_encode($output)); 
mysql_close();

The logcat:org.json.JSONException: Value

What should i do? The Php code runs on web page without any errors but why got an error in this part? In logcat also has: ( ! ) Notice: Undefined variable: user_id in C:\\wamp\\www\\getProducts.php on line 27

Logcat!

<br /><font size='1'><table class='xdebug-error xe-notice' dir='ltr' border='1' cellspacing='0'    cellpadding='1'><tr><th align='left' bgcolor='#f57900' colspan="5"><span style='background-color: #cc0000; color: #fce94f; font-size: x-large;'>( ! )</span> Notice: Undefined variable: user_id in C:\wamp\www\getProducts.php on line <i>27</i></th></tr><tr><th align='left' bgcolor='#e9b96e' colspan='5'>Call Stack</th></tr>... this kind of code appears in logcat then org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONArray at org.json.JSON.typeMismatch(JSON.java:107)

Answer 1

You do not return a valid JSON, seems that you are returning a HTML snippet with <br> have a look at your returning source of your website.

What should i do? The Php code runs on web page without any errors but why got an error in this part? In logcat also has: ( ! ) Notice: Undefined variable: user_id in C:\\wamp\\www\\getProducts.php on line 27

So the user_id is not defined! Just look at your GET/POST Parameters or show us your relevant php code.

Please use echo json_encode($output); not print .

And what are you doing if there is no returning row from your first Query. Try this.

$sql=mysql_query("SELECT user_id FROM users WHERE user_name='".$user_name."' AND user_pass='".$password."' "); 
if(mysql_num_rows($sql) == 0){
   echo "USERID CANNOT BE FOUND";
}
while($row=mysql_fetch_array($sql)) { 
  echo "USERID FOUND" .$row["user_id"] ;
  $user_id=$row["user_id"]; 
} 

It seems that user_name could not be found. So just debug it there by trying to echo your user_id after the while. If you do not get the UserID FOUND echo the user_name and password do not exist.

Edit: As your logact says, there is no Entry for the username you are passing so the $user_id will never be filled.

http://192.168.2.245/getProducts.php?login=1&user_name=UserName&password=Password

You are passing user_name = UserName and password = Password. Does this entry really exist in your Database?

Also for better performance have a look at MySQL LETF JOIN and MySQL subqueries.