Need help fetching an array from a prepared statment - php

Question

I am new to using prepared statement with PHP. I am trying to get the value of "full_name"... So far I am stuck over here. Can anyone please help figure this out? Thanks!
if($db->connect_error){
echo "Connection Error";
}
$id = 834;
$stmnt = $db->prepare("SELECT * FROM checkout_page where id = ?");
$stmnt->bind_param("i", $id);
if (!$stmnt->execute()) {
echo "Execute failed: (" . $stmnt->errno . ") " . $stmnt->error;
}
$row = $stmnt->fetch();

Answer 1

You need to use bind_result to bind variables to the columns you want. Then each time you call fetch(), those variables will be updated with the next row's values. fetch() with mysqli does not return you the row/result.
This means you cannot use SELECT *. You need to specify which fields you want.
if($db->connect_error){
echo "Connection Error";
}
$id = 834;
$stmnt = $db->prepare("SELECT full_name FROM checkout_page where id = ?");
$stmnt->bind_param("i", $id);
if (!$stmnt->execute()) {
echo "Execute failed: (" . $stmnt->errno . ") " . $stmnt->error;
}
$stmnt->bind_result($full_name);
$stmnt->fetch();
echo $full_name;
Or, if you have the mysqlnd driver installed, you can use get_result() to get a result set just like if you had ran a normal query, not a prepared statement.
if($db->connect_error){
echo "Connection Error";
}
$id = 834;
$stmnt = $db->prepare("SELECT * FROM checkout_page where id = ?");
$stmnt->bind_param("i", $id);
if (!$stmnt->execute()) {
echo "Execute failed: (" . $stmnt->errno . ") " . $stmnt->error;
}
$result = $stmnt->get_result();
$row = $result->fetch_assoc();
echo $row['full_name'];