I get:

Undefined variable: password
Undefined variable: host
Undefined variable: user

I am just curious why I am getting such notices, although the variable has been defined in the private section of the class.

Can't I use private data member of the class in the member function (because that would defeat the whole concept of OOP)?

The php file is:

class data_base //helps handling permissins
    {
    private $host;
    private $user;
    private $password;

    public function feed_data($hst, $usr, $pwd)
        {
            $host=$hst;
            $user=$usr;
            $password=$pwd;
        }
    public function get_data()
        {
            $info=array("host"=>" ", "user"=>" ", "password"=>" ");
            $info['host']=$host;
            $info['user']=$user;
            $info['password']=$password;    
            return $info;
        }
      }

   $user1=new data_base; 
   $user2=new data_base;

   $user1->feed_data("localhost", "root", ""); //enter details for user 1 here
   $user2->feed_data("", "", ""); //enter details for user 2 here

   $perm_add=$user1->get_data();
   $perm_view=$user2->get_data();

• php
• warnings

In PHP you must call a property as a property

$this->host;
// instead of
$host;

Unlike for example in java $host is always a local variable and thus undefined here.

As a sidenote: You can write

$info=array("host"=>" ", "user"=>" ", "password"=>" ");
$info['host']=$host;
$info['user']=$user;
$info['password']=$password;    
return $info;

as

return array(
    'host'     => $this->host,
    'user'     => $this->user,
    'password' => $this->password
);

It's short and imo more reable (No need for a temporary variable)

In PHP, to access instance variables, you need to use $this->varname.
Just $varname is always a variable local to the method.