如何在mysql_query php中选择一个表

hi to all m new to OOP PHP. i have following piece of code here in the following:

test.php

<?php
class test 
{
    private $dbhost;
    private $dbname;
    private $user;
    private $pass;
    public function __construct()
    {
        $this->dbhost = 'localhost';
        $this->dbname = 'XXXX';
        $this->user = 'root';
        $this->pass = '';
    }
    public function open($obj)
    {
        $con = mysql_connect($this->dbhost,$this->user,$this->pass) or die(mysql_error());
        $db = mysql_select_db($this->dbname)or die(mysql_error());
        $query = mysql_query("SELECT * FROM '".$obj."'") or die(mysql_error());

    }

}

?>

in the other file named test2.php i have the following piece of code:

<?php
    require_once('testClass.php');
    $obj = new test();
    $obj2 = 'user';
    $obj->open($obj2);
?>

and i got the following error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''user'' at line 1

thnx in advance for help

$obj should be tablename and you doesn't need to put single quote around table name.

mysql_query("SELECT * FROM ".$obj) or die(mysql_error());

Edit:

 while($row = mysql_fetch_array($query)){
    echo $row['userId'];
 } 

Suggestion: Your query must be affected by mysql injection attack. It is really good to use PDO or Mysqli lib.

$sql = mysql_query("SELECT * FROM your_table");
$row = mysql_num_row($sql);
if($row > 0){
//do something
}else{
echo "No records found";
}