Why is the result null when var_dump () is used to print the mysqli object?

The Code is as follows: {code ...}

The Code is as follows:

$mysqli=new mysqli("localhost:3307","root","","test");var_dump($mysqli);print_r($mysqli);

Reply content:

The Code is as follows:

$mysqli=new mysqli("localhost:3307","root","","test");var_dump($mysqli);print_r($mysqli);

$ Mysqli = new mysqli ("localhost: 3306", "root", "", "test ");
Var_export ($ mysqli );
Echo"
";
Var_dump ($ mysqli );
Echo"
";
Print_r ($ mysqli );

The returned result is mysqli ::__ set_state (array ('Affected _ rows '=> NULL, 'client _ info' => NULL, 'client _ version' => NULL, 'connect _ errno' => NULL, 'connect _ error' => NULL, 'errno' => NULL, 'error' => NULL, 'Field _ count' => NULL, 'host _ info' => NULL, 'info' => NULL, 'insert _ id' => NULL, 'server _ info' => NULL, 'server _ version' => NULL, 'STAT' => NULL, 'sqlstate' => NULL, 'Protocol _ version' => NULL, 'thread _ id' => NULL, 'Warning _ count' => NULL ,))

Checked var_export.
Var_export must return valid php code, that is, the Code returned by var_export can be directly assigned a variable as a php code. And this variable will get the same type value as var_export.
However, when the variable type is resource, it cannot be simply copied. Therefore, when the var_export variable is resource type, var_export returns NULL.

A similar problem may be caused by the landlord.