显示基于URL PDO的结果

Question

I am fetching results from database using PDO with url query string

url: index.php?bType=doctor&loc=asia|india

CODE:

$bind = array();
$str=explode('|',$_GET[loc]); 
  foreach ($str as $loc) 
{
    $bloc[] = $loc;
}   $loca = implode("','", $bloc); 
 $btype=$_GET['bType'];

$sqlsb = "SELECT * FROM t_business WHERE 1=1";
if(isset($_GET['bType'])){ $sqlsb .=" AND type like :btype"; $bind['btype']='%'. $btype .'%';}

if(isset($_GET['loc'])){ $sqlsb .=" AND location IN (:loca)"; $bind['loca']="'$loca'";}

$qsb = $db->prepare($sqlsb);
$qsb->execute($bind);

---

above code fetches nothing..

print_r($bind) shows Array ( [btype] => doctor [loca] => asia','india )

If i do it without using prepared it works.

I am writing this code for refine search filters.

Thanks

UPDATE

AS CoursesWeb anwserd i did changes

if(isset($_GET['bType'])){ $sqlsb .=" AND type like :btype"; $bind['btype']='%'. $btype .'%';}
if(isset($_GET['loc'])){ $sqlsb .=" AND location IN (:loca)";  $bind['loca']="'$loca'";}

it works for :btype but not with :loca

print_r($bind) shows array ( [btype] => %doctor% [loca] => 'asia','india' )

Need to do something better with loc than using implode..think so

Answer 1

for prepared values used in LIKE statement, the "?" or "%" characters must be added together with the value, not in sql statement.

$btype = '%'. $btype .'%';
$sqlsb = "SELECT * FROM t_business WHERE type LIKE :btype";