Why is the result null when printing mysqli objects with Var_dump ()?

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 the next var_export.
Var_export must return a valid PHP code, that is, the code returned by Var_export can be directly assigned to a variable as PHP code. And this variable gets the same type of value as the Var_export.
However, when the variable type is resource, it cannot be copied in simple copy, so var_export returns NULL when the Var_export variable is resource type.

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