Get JSON data FROM a PHP/MySQL query into a html tag using jquery

probably safer to do like this:

echo json_encode(array("user" => $arr));

on the other end you would receive an object which, I would suggest iterating like this:

var k;

for (k in data.user){
     $("#getname").append($("<span></span>").html(data.user[k].name));
} 

Given that you are fetching information for one user only, following I would suggest

$id = (int) $_GET["id"]; // or wherever you get it from.
if ($r = $db->mysql_fetch_assoc()){
    $response = array(
        "name" => $r["name"];
    );
    echo json_encode($response);
} else {
    echo json_encode(array("error" => "Could not get name for user " . $id));
}

Then, on front-end, all you need to do is:

if (typeof(data.name) != "undefined"){
    $("#getname").html(data.name);
} else if (typeof(data.error) != "undefined"){
    $("#getname").html(data.error); //or handle otherwise
}

Your JSON is malformed. You are appending a bunch of objects {.1.}{.2.}{.3.}. Instead, try {"users":[{.1.},{.2.},{.3.}]}.

In PHP you'll do something like this (note that I've changed the response type to JSON-P rather than JSON by adding a callback parameter):

public function User($id) {
    $users = array();

    $this->connect_db_web($conn);
    $sql = mysql_query("SELECT * FROM users WHERE id='".$id."'");
    while ($values = mysql_fetch_array($sql)) {
        $users[] = array(
          'id'=>$values['idUsers'],
          'name'=>$values['name']
          // etc.
        );
    }
    $obj['users'] = $users;

    $callback = (empty($_GET["callback"])) ? 'callback' : $_GET["callback"];
    echo $callback . '(' . json_encode($obj) . ');';
}

Then you'll be able to do:

$.getJSON("user.php?callback=",function(data){
  $.each(data.users, function(i,user){
    $('#getname').append(user.name);
  });
}); 

You've misinterpreted your JSON structure. You're appending your DB rows to an array, and embedding that inside an object. If you'd do a console.log(user) inside your .getJSON call, you'd see you'll have to do:

user[0].name

instead. As well, your code assumes that the user ID exists, and returns data regardless of how many, or how few, rows there actually are in the result set. At minimum your JS code code should check users.length to see if there ARE are any rows to begin with. Beyond that, unless you're doing it in another section of code somewhere, that $id value is probably coming from the web page, which means your query is vulnerable to SQL injection attacks.

OK got it, was a php code error and JSON structre as marc said, here I´m gonna post what finally I had PHP Class

public function User() {
$users = array();
$this->connect($conn);
$sql = mysql_query("SELECT * FROM users WHERE id='1'");
    $values = mysql_fetch_array($sql);
        $users[] = array(
            'id'=>$values['id'],
            'name'=>$values['name'],
            'name2'=>$values['name2'],
            'lname'=>$values['lname'],
            ...//rest of values
            );

    echo json_encode($users);
}

PHP module to get class

include"class.php";
$user = new Users();
$user->User();

Now how did I got the values using JQuery

$.getJSON('user.php', function(data){
        $('wherever_you_want_to_point_at').text(data[0].name);
    });

Hope it helps someone, Thanks again guys, very very helpful Take care you all