Why does a new mysqli object return false if it did work?

Asked Mar 04 '14 at 19:25

Active Mar 04 '14 at 19:25

Viewed 78 times

maybe you can enlighten me. Is there a good explanation why the mysqli object returns false if it did work and true if it didn't?

Code:

$connection= @new mysqli($ip, $user, $password);

if (!$connection)
{
    $connection->select_db($db);
    $connection->set_charset("utf8");
}

Why is this correct and not if ($connection)?

asked Mar 04 '14 at 19:25

CodeShark

It isn't correct. It should be `if ($connection)`. – Barmar Mar 04 '14 at 19:27
4

Do you understand the meaning of the @ or ! ? Read the docs – Ronni Skansing Mar 04 '14 at 19:28
I guess this is the correct way to check if you've connected successfully: http://stackoverflow.com/a/3678682/679227 – Haralan Dobrev Mar 04 '14 at 19:31

0 Answers0