Jsonp retreive data problem!

Hello,
I have set everything to sent asynchronous request to php file where it retrieve a json encoded details and things works fine but i get this error when i edit this


      This works fine !!
      
      $arr = array ('item1'=>"1",'item2'=>"alo",'item3'=>"We love jQuery4u");

      but when i tried to assign a variable
      for example

//details from database
$user = $row['details'];
$arr = array ('item1'=>"1",'item2'=>$user,'item3'=>"We love jQuery4u");

i got in the jquery response null related to item2!

so i figure out when assigning to the array a variable instead of constant it give me error so how can this be fixed
jaylab2Asked:
Who is Participating?
 
RobConnect With a Mentor Owner (Aidellio)Commented:
http://www.egnyt.co/alain/?user=medialablebanon@gmail.com&psw=12345678 shows null.

As i've said previously, please change your code from:
header('Content-type: application/javascript; charset=utf-8');
echo "/**/typeof ".$cb."==='function' && ".$cb."(".json_encode($arr).")";

Open in new window

to:
header('Content-type: application/json; charset=utf-8');
echo json_encode($arr);

Open in new window

0
 
Ray PaseurCommented:
Please post the complete text of the JSON string that is returned from the PHP service.  Once we see the data it should be easy to show you how to process it.  Thanks.
0
 
RobOwner (Aidellio)Commented:
Your error is that your php $user variable hasn't been set.  It's sent that way to the browser (JavaScript)

Will need to see more of your php code, specifically the score between where you sty the user variable and the point you send the data back to the browser
0
 
jaylab2Author Commented:
Here it is

function login($user,$password){

	$u = $user;

	$first = "SELECT * FROM customer_entity where email='".$user."'";
	
	$sql = mysql_query($first);

	//$row = mysql_fetch_array($sql);
	
	$check1 = mysql_num_rows($sql);
	
	//get the customer id
	if ($check1 > 0 ){
	
		
		$code = mysql_fetch_array($sql);
		
		$userid = $code['entity_id'];
		
		//GET password hash
		$second = "SELECT * FROM customer_entity_varchar where entity_id='".$userid."' ORDER BY value_id ASC LIMIT 2,4";
		
		$sq = mysql_query($second);
		
		$test = mysql_fetch_array($sq);
		
		$pswhash = $test['value'];
		
		//now split 
		$hash = explode(":",$pswhash);
		
		$salt= $hash[1];
		
		$pass = md5($salt.$password);
		
		//return $pass."<br>".$hash[0];
		
		//GET NOW THE USERNAME 
		$third = "SELECT * FROM customer_entity_varchar where entity_id='".$userid."' ORDER BY value_id ASC LIMIT 0,4";
		
		$sq1 = mysql_query($third);
		
		$fname1 = mysql_fetch_array($sq1);
		
		$fname = "'".$fname1['value']."'";
		
		//GET NOW THE LASTNAME
		$fourth = "SELECT * FROM customer_entity_varchar where entity_id='".$userid."' ORDER BY value_id ASC LIMIT 1,4";
		
		$sq2 = mysql_query($fourth);
		
		$lname1 = mysql_fetch_array($sq2);
		
		$lname = $lname1['value'];
		
		if ($pass = $hash[0]) $auth =  "1";
		else $auth = "2";
		
	
	
	}else $auth = "3";

//return array ('item1'=>$auth,'item2'=>$user,'item3'=>$fname,'item4'=>$lname);
return array ('item1'=>'1','item2'=>$u,'item3'=>$fname,'item4'=>'slim');
}

Open in new window



//HERE GOES THE INCLUDES FOR CONNECTION 
include 'includes/dbconnect.php'; 
include 'includes/sql.php';
//END DATAS CONNECT

// where $_GET['callback'] = 'randomFn123'
$cb = $_GET['callback'];

//$arr = array ('item1'=>"I love jquery4u",'item2'=>"You love jQuery4u",'item3'=>"We love jQuery4u");

//START LOGIN
if (isset($_REQUEST['user'])){

	$arr = login($_REQUEST['user'],$_REQUEST['psw']);
	
	//$arr = array ('item1'=>"1",'item2'=>"medialablebanon@gmail.com",'item3'=>"We love jQuery4u");
	
}


if (preg_match('/\W/', $cb)) {
  // if $_GET['callback'] contains a non-word character,
  // this could be an XSS attack.
  header('HTTP/1.1 400 Bad Request');
  exit();
}


header('Content-type: application/javascript; charset=utf-8');
echo "/**/typeof ".$cb."==='function' && ".$cb."(".json_encode($arr).")";

Open in new window

0
 
RobOwner (Aidellio)Commented:
Your header should be "application/json" not JavaScript

Then follow that with

echo json_encode($arr);
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.