使用AJAX从php接收数据

I am working on a function, where if a user selects a value from a drop down list, then this calls an AJAX function, which then should get me a list of Names and their email addresses

$.ajax({
                type:'POST',
                url:'get_distusers.php',
                data:'dist_id='+dist_id+'&study_id='+study_id,
                success: function(data){

                    $('#dist_list2').html(data);
                    $('#dist_list').val(data);

                },
                    error : function(XMLHttpRequest, textStatus, errorThrown) {
                        alert("There was an error."+textStatus+" - " + XMLHttpRequest.status);
                    }
            });

Then the other page has this to send back the list of names and emails

unset($contact);
if ($count_du>0) {
    while($row_du = mysqli_fetch_array($result_du)) {
        if(!empty($contacts)) 
            $contacts.="<br>, ";

            $contact['name']=$row_du[0];
            $contact['email']=$row_du[1];
    }
}

$contacts[]=$contact;
//$contacts=array($contact_email,$contact_name);

//echo $sql_patrand;
echo json_encode($contacts);

However, when I try to retrieve the contact list, I either get the word object or nothing.

Can someone tell me what I am doing wrong?

You have return echo json_encode($contacts); json format from the your php code, but your ajax not understand json format because you have not set dataType: 'json' param, Please check the below mention code, it will defiantly helpful, any query comment inside answer....

$.ajax({
    type:'POST',
    url:'get_distusers.php',
    data:'dist_id='+dist_id+'&study_id='+study_id,
    dataType: 'json', //NOTE:Your missing point - update first
    success: function(data){
        console.log(data); //Print your data, so it will help you to how manage
        $('#dist_list2').html(data);
        $('#dist_list').val(data);

    },
        error : function(XMLHttpRequest, textStatus, errorThrown) {
            alert("There was an error."+textStatus+" - " + XMLHttpRequest.status);
        }
});