CODEWITHSUNDEEP
phpsqlprepared-statement
Stephen
Stephen

Reputation: 390

Update to prepared SQL statement not returning results

I'm having trouble updating a site. With security in mind, I am trying to rewrite the SQL statements using PDO prepared. It's my preferred choice generally.

The site I'm working on has this query, returning results via json to a search box. It works ...

$sql = "SELECT * FROM stock_c_ranges WHERE deleted = 'no' AND current_status = 'current'";
$result = mysqli_query($conn, $sql);
    $results_list = array();
while($row = mysqli_fetch_array($result)){
    $results_list[] = $row['manufacturer_name'].' ID:'.$row['id'];
}

echo json_encode($results_list);

I've re-written using prepared statements ...

$range = "SELECT * FROM stock_c_ranges WHERE deleted= :deleted AND current_status= :cstatus";
$deleted='no';
$c_status='current';
$result_lists=array();
$stmt=$pd_con->prepare($range);
$stmt->bindParam(':deleted',$deleted,PDO::PARAM_STR);
$stmt->bindParam(':cstatus',$c_status,PDO::PARAM_STR);
$stmt->execute;
while($row=$stmt->fetch(PDO::FETCH_ASSOC)){
    $results_list[] = $row['manufacturer_name']. 'ID:'.$row['id'];
}
 echo json_encode($results_list);

..... this doesn't

I've either made a glaring syntax error that I'm just blind to after looking at it for so long, or there is something about using PDO and JSON/AJAX that I'm not aware of stopping it functioning.

Upvotes: 2

Views: 33

Answers (1)

Stephen
Stephen

Reputation: 390

Apologies, writing it on here has highlighted the glaringly obvious ...

$stmt->execute;

Should have been ...

$stmt->execute();

Upvotes: 1

Related Questions