mysqli doesn't return instance with die() - php

Question

I have function :
function dbConnect($usertype, $connectionType = 'mysqli') {
// some code hare
return new mysqli($host, $user, $pwd, $db) or die ('Cannot open database');
}
but when I try this:
$conn = dbConnect('read');
$result = $conn->query('SELECT * FROM images');
function dosn't return anything and it say :
Fatal error: Call to a member function query() on a non-object in
C:\xampp\htdocs\phpsols\mysql\mysqli.php on line 10
but it works this way(without die())
return new mysqli($host, $user, $pwd, $db);

Answer 1

The [..] or die() construct leads to funny behaviour in conjunction with the return statement: The whole thing is interpreted as a boolean expression.
And because new mysqli will never be false, the "die" is never processed, and thus, your function returns true instead of a newly created instance of mysqli.
If you still would like to use or die(), do this:
$result = new mysqli($host, $user, $pwd, $db) ;
if (!$result) die ("Can't open database.");
return $result;

Answer 2

You can try this:
function dbConnect($usertype, $connectionType = 'mysqli') {
// some code hare
try {
return new mysqli($host, $user, $pwd, $db);
} catch(Exception $e) {
die ('Cannot open database');
}
}

Answer 3

aside from this funny one-liner problem, your idea of error handling is all wrong. Sadly, it was copy-pasted to other answer as well.
Here is the right code
$mysqli = new mysqli($host, $user, $pwd, $db);
if ( mysqli_connect_errno() )
{
throw new Exception(mysqli_connect_errno()." ".mysqli_connect_error());
}
return $mysqli;
Honestly, this die ('something'); thing is a childhood disease of PHP. Time to grow up a little.