Using pdo php in my class

Question

I use pdo php in my class

<?php 
class User 
 {
   private $db;
   public function __construct($host, $user, $pass, $db)
   {
    try
     {
      $this->db = new PDO("mysql:host=$host;dbname=$db", $user, $pass); 

     }
    catch (Exception $e)
     {
      die('Error : ' . $e->getMessage());
     }
   } 
  public function modUser($uid, $email)
   {
    $query = $this->db->exec("UPDATE Users SET  Email =:email  WHERE Id_User =:uid");
         $query->bindValue(':email', $email, PDO::PARAM_STR);   
   }
}
?>

When i use the class user like this:

$user = new User('localhost','tester','0000','Agency);

It works well,but the problem is when use config file to connect database and write

$user = new User($host, $user, $pass, $db);

I get an error:

<b>Warning</b>:  PDO::__construct() expects parameter 2 to be string, object given in <b>C:\wamp\www\new_template-latest\new_template\classes\class.User.php</b> on line <b>13</b><br />
<br />
<b>Fatal error</b>:  Call to a member function exec() on a non-object in <b>C:\wamp\www\new_template-latest\new_template\classes\class.User.php</b> on line <b>140</b><br />

Can you please include the config file? PHP may be loading it from the config file and typing it as an integer instead of a string. — Sean
– Sean, Commented Jul 8, 2011 at 15:10
It appears from his code that a new User object is stored in $user — Charles Sprayberry
– Charles Sprayberry, Commented Jul 8, 2011 at 15:10
When using the config file, if you add the following line what is output? var_dump($pass); exit; — David Snabel-Caunt
– David Snabel-Caunt, Commented Jul 8, 2011 at 15:11

Charles Sprayberry · Accepted Answer · 2011-07-08 15:10:26Z

5

You are trying to assign the new User object to the PDO parameters. You need to choose a new variable name for your User object or a new name for your $user database configuration variable.

answered Jul 8, 2011 at 15:10

Charles Sprayberry

7,8823 gold badges43 silver badges52 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

bilal Over a year ago

Exactly the problem is in variables name ,i change variable name it's work .thank you for all

Jack Murdoch · Accepted Answer · 2011-07-08 15:12:34Z

2

You are passing the $user variable straight into the constructor, which is now a new User object. You should store the username string in a variable with a different name.

answered Jul 8, 2011 at 15:12

Jack Murdoch

2,8941 gold badge20 silver badges27 bronze badges

Collectives™ on Stack Overflow

Using pdo php in my class

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related