I'm trying to make a function that receive a query (sql) and a parameter (array) but I receive this error:
PHP Warning: Parameter 2 to mysqli_stmt::bind_param() expected to be a reference, value given
My code is:
function dbRead($query, $param) {
$mysqli = new mysqli(DB::READ_HOST, DB::READ_USER, DB::READ_PASS, DB::NAME);
// Check that connection was successful.
if ($mysqli->connect_error) {
$result = "Connection error";
} else {
// Check that $conn creation succeeded
if ($conn = $mysqli->prepare($query)) {
call_user_func_array(array($conn, 'bind_param'), $param);
$conn->execute();
$result = $conn->get_result();
$result = $result->fetch_array();
$conn->close();
} else {
$result = "Prepare failed";
}
}
$mysqli->close();
return $result;
}
$test = dbRead('SELECT * FROM user WHERE id=? and email=?', array(123,'example@example.com'))
And if my function code is
function dbRead($query, $param) {
$mysqli = new mysqli(DB::READ_HOST, DB::READ_USER, DB::READ_PASS, DB::NAME);
// Check that connection was successful.
if ($mysqli->connect_error) {
$result = "Connection error";
} else {
// Check that $conn creation succeeded
if ($conn = $mysqli->prepare($query)) {
$ref = array();
foreach ($param as $key => $value) {
$ref[$key] = &$param[$key];
}
call_user_func_array(array($conn, 'bind_param'), $ref);
$conn->execute();
$result = $conn->get_result();
$result = $result->fetch_array();
$conn->close();
} else {
$result = "Prepare failed";
}
}
$mysqli->close();
return $result;
}
I receive this error
PHP Warning: mysqli_stmt::bind_param(): Number of elements in type definition string doesn't match number of bind variables
My PHP version is 5.4.36