PHP - warning - Undefined property: stdClass - fix?

if(isset($response->records))
    print "we've got records!";

Every time you do $array[] = it inserts a new element in the end of an array. What you need to do is:

class MaterialType {

public $id;
public $name;


function getAllMaterialType() {

    $query = "SELECT * FROM material_types";
    $result = mysql_query($query);

    $arr = array();
    while($row = mysql_fetch_array($result)) {
        $mat = new MaterialType();
        $mat->id = $row['m_type_id'];
        $mat->name = $row['m_type_name'];
        $arr[] = $mat;
    }

    return $arr;
}

}

edit: it's currently 2016-09-24, and PHP 5.4 has been released 2012-03-01, and support has ended 2015-09-01. Still, this answer seems to gain upvotes. If you're still using PHP < 5.4, your are creating a security risk and endagering your project. If you have no compelling reasons to stay at <5.4, or even already use version >= 5.4, do not use this answer, and just use PHP>= 5.4 (or, you know, a recent one) and implement the JsonSerializable interface

You would define a function, for instance named getJsonData();, which would return either an array, stdClass object, or some other object with visible parameters rather then private/protected ones, and do a json_encode($data->getJsonData());. In essence, implement the function from 5.4, but call it by hand.

Something like this would work, as get_object_vars() is called from inside the class, having access to private/protected variables:

function getJsonData(){
    $var = get_object_vars($this);
    foreach ($var as &$value) {
        if (is_object($value) && method_exists($value,'getJsonData')) {
            $value = $value->getJsonData();
        }
    }
    return $var;
}