Solved

Broken function after implementing PDO

Posted on 2014-03-13
5
352 Views
Last Modified: 2014-03-13
I have decided to upgrade my code to keep uptodate with mysql.
I have a function which worked fine before but now doesnt...

Firebug is giving this error:

GET http://localhost/stinteractive/touch/library/process_user.php?action=findUsers&term=ste      500 Internal Server Error      15ms      jquery....min.js (line 6)
"NetworkError: 500 Internal Server Error - http://localhost/stinteractive/touch/library/process_user.php?action=findUsers&term=ste"

Could someone help!

OLD
function findUsers()
{
//$text = mysql_real_escape_string($_GET['term']);
$text = $_GET['term'];

$query = "SELECT userID, userFirstname, userSurname FROM user WHERE CONCAT(userFirstname, ' ', userSurname) LIKE '%$text%' OR userSurname LIKE '%$text%' ORDER BY userSurname ASC";
$result = mysql_query($query);
$json = '[';
$first = true;


while($row = mysql_fetch_array($result)) {
    if (!$first) {
		$json .=  ',';
	} else {
		$first = false;
	}
    
	$json .= '{"userID":"'.$row['userID'].'",';
	$json .= '"value":"'.$row['userFirstname'] .' '.$row['userSurname'].'"}';
}
$json .= ']';
echo $json;
}

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NEW (and doesnt work)
function findUsers()
{
	//$text = mysql_real_escape_string($_GET['term']);
	$text = $_GET['term'];

	try
	{
		$sql = $db->prepare('SELECT userID, userFirstname, userSurname
			FROM user
			WHERE CONCAT(userFirstname, " ", userSurname) LIKE ?
			OR userSurname LIKE ?
			ORDER BY userSurname ASC');
		$sql->execute(array("%".$text."%","%".$text."%"));

		$sql->setFetchMode(PDO::FETCH_ASSOC);
		$data = $sql->fetchAll();
	}

	catch(PDOException $e)
	{
		echo $e->getMessage();
		die;
	}

	$json = '[';
	$first = true;

	foreach ($data as $row) {

	    if (!$first) {
			$json .=  ',';
		} else {
			$first = false;
		}
	    
		$json .= '{"userID":"'.$row['userID'].'",';
		$json .= '"value":"'.$row['userFirstname'] .' '.$row['userSurname'].'"}';
	}
	$json .= ']';
	echo $json;

}

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0
Comment
Question by:sjtinsley83
5 Comments
 
LVL 108

Accepted Solution

by:
Ray Paseur earned 500 total points
ID: 39927473
$db is undefined in the scope of the findusers() function.  Maybe make it a global?
0
 
LVL 34

Expert Comment

by:Dan Craciun
ID: 39927481
I think you forgot the ''

$sql = $db->prepare('SELECT userID, userFirstname, userSurname
			FROM user
			WHERE CONCAT(userFirstname, " ", userSurname) LIKE '?'
			OR userSurname LIKE '?'
			ORDER BY userSurname ASC');
		$sql->execute(array("%".$text."%","%".$text."%"))

Open in new window

HTH,
Dan
0
 
LVL 58

Expert Comment

by:Gary
ID: 39927489
...Or pass it to your function

@Dan
They are placeholders, you don't encapsulate them.
0
 

Author Comment

by:sjtinsley83
ID: 39927608
This
global $db;
in the function fixed it!
0
 
LVL 58

Expert Comment

by:Gary
ID: 39927616
Then you should accept Rays comment and not mine
Re-opening.
0

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