multiple mysqli objects overwrite each other

Question

If I create two parallel connections to two servers:

$gw03 = new mysqli('gw03.example', 'user', 'pass', 'db');
$gw04 = new mysqli('gw04.example', 'user', 'pass', 'db');

if (!$gw03->connect_errno) {
  ...
} else if (!$gw04->connect_errno) {
  ...
} else {
  echo "Failed to connect to gw03: (" . $gw03->connect_errno . ") " . $gw03->connect_error . PHP_EOL;
  echo "Failed to connect to gw04: (" . $gw04->connect_errno . ") " . $gw04->connect_error . PHP_EOL;
}

If gw03 is available but gw04 is not, the following is the result

Failed to connect to gw03: (2002) No route to host
Failed to connect to gw04: (2002) No route to host

The failure to connect to $gw04 seems to overwrite $gw03. Aren't $gw03 and $gw04 separate objects? What's going on here?

Answer 1

Unfortunately, mysqli suffers from bad design. The properties connect_errno and connect_error are not really properties. They are just shortcuts to global variables. When the second connection fails the global variable is overwritten internally.

This is a very good reason to stop checking for mysqli errors manually. When an error happens your code should throw an error instead of a warning or failing silently. This error should be properly handled by your application and logged in a secure place on the server. Don't try-catch or echo the error message to the user!

To enable mysqli error reporting you should add the following line before any new mysqli()/mysqli_connect():

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

This setting will tell mysqli to throw an exception as soon as an error is encountered. No more manual checking and no more problems with global variables overwriting themselves.

Answer 2

They are separate connections but the connect_errno method "returns the error code from last connect call".

https://www.php.net/manual/en/mysqli.connect-errno.php

You will want to check for a connection error right after each connection attempt.

$gw03 = new mysqli('gw03.example', 'user', 'pass', 'db');
# check $gw03 for connection error before attempting next connection
$gw04 = new mysqli('gw04.example', 'user', 'pass', 'db');