Is this correct mysqli - mysql

Question

Is this correct, I know you aren't supposed to use mysql anymore. So I just added i to the end, this was code I got from the internet, and I am new to mysql so I'm not quite sure what I am doing. I just want to make sure this is correct usage.
$sql = "SELECT lat, lon from zipcodes_2011 WHERE zipcode='$zip'";
$r = mysqli_query($con, $sql);
if (!$r) {
$this->last_error = mysqli_error();
return false;
} else {
$row = mysqli_fetch_array($r);
mysqli_free_result($r);
return $row;
}

Answer 1

That is incorrect.
Looking at the documentation for mysqli_error: it states that mysqli $link is not optional. You must pass the database link to it.
$this->last_error = mysqli_error($con);
I would also highly recommend preparing this.
$sql = "SELECT `lat`,`lon` FROM `zipcodes_2011` WHERE `zipcode` = ?";
$stmt = mysqli_stmt_init($link);
$error = mysqli_stmt_prepare($stmt, $sql);
if ($error === false) {
$this->last_error = mysqli_error($con);
return false;
}
$error = mysqli_stmt_bind_param($stmt, "s", $zip);
if ($error === false) {
$this->last_error = mysqli_error($con);
return false;
}
$error = mysqli_stmt_execute($stmt);
if ($error === false) {
$this->last_error = mysqli_error($con);
return false;
}
$error = mysqli_stmt_bind_result($stmt, $row);
if ($error === false) {
$this->last_error = mysqli_error($con);
return false;
}
$error = mysqli_stmt_fetch($stmt);
if ($error === false) {
$this->last_error = mysqli_error($con);
return false;
}
return $row;