Question

I have the following mySQL query:

$content = mysqli_query($dblink, "SELECT * FROM passwordrecovery WHERE code = '$formcode'");

            while($row = mysqli_fetch_array($content)) {
                $db_username = $row['username'];
                $db_email = $row['email'];
                $db_code = $row['code'];
            }

I would like to add an IF statement somewhere here to check whether the query is successful but I am not sure where.

Basically, I have a form where a user can insert a code (sent by email) to retrieve their password. But they might make a mistake and enter a code that doesn’t exist in the database and I would like to account for such a situation and throw back an error. I am just not sure how to do it?

Answer 1

On query failure, $content will be FALSE rather than a result resource. Check that it is not false. If they entered a non-existent code, you would get no rows back, so check for at least one row via mysqli_num_rows().

$content = mysqli_query($dblink, "SELECT * FROM passwordrecovery WHERE code = '$formcode'");

if ($content) {
    // Check for at least one row returned
  if (mysqli_num_rows($content) < 1) {
    // Invalid lookup, show error to user
  }
  // Got a row back, so query was valid
  else {
    while($row = mysqli_fetch_array($content)) {
      $db_username = $row['username'];
      $db_email = $row['email'];
      $db_code = $row['code'];
    }
  } 
}
else {
  // Query failure - display error to user
  // and check mysqli_error() internally for debugging
}

We assume you have already properly escaped $formcode via mysqli_real_escape_string($formcode).