Question

Hi guys I´m new at stackoverflow and also new at Jquery

Well hope I can make myself understandable. Here is what I want: I have made a query to my MySQL db, using a class with PHP

public function User($id) {
    $this->connect_db_web($conn);
    $sql = mysql_query("SELECT * FROM users WHERE id='".$id."'");
    while ($values = mysql_fetch_array($sql)) {
        $arr[]=array(
        'id'=>$values['idUsers'],
        'name'=>$values['name'],
        'name2'=>$values['name2'],
        'lname'=>$values['lname'],
        'lname2'=>$values['lname2'],
        'email'=>$values['email'],
        'phone'=>$values['phone'],
        'address'=>$values['address'],
        'bday'=>$values['bday'],
        'password'=>$values['password']
        );
    }
    echo '{"user":'.json_encode($arr).'}';
}

Then I have a php code where I call this function

$name = $user->User($id);

I think this works ok (if I´m wrong please help). Now what I´m really trying to do is getting the values from the JSON array into specific divs, example:

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

And inside my HML i Have a <p id="getname"></p>wich is the tag I want the value to be displayed

But no value is displayed, why?, what am I doing wrong?

Thanks for the help I apreciate it

Answer 1

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
}