Php 在ajax成功函数中返回两个变量

如何在一个函数中同时返回地块号和注释数？我有两个单独的sql字符串返回不同的结果。我希望能够返回apn编号，并将值放入标签中，以及有多少条评论的计数。这可能吗？我该怎么做

Jquery：

$.ajax({
    url: "classes/get-apn-count-comments.php?parcel_id=" + parcel_id,
    type: "GET",
    data: { parcel_id : parcel_id },
    dataType: 'json',
    error: function(SMLHttpRequest, textStatus, errorThrown){
        alert("An error has occurred making the request: " + errorThrown);
    },
    success: function(data2){
        //do stuff here on success
        //$('#ParcelNumber').html(data[0]["apn"]);
        $('#ViewComments').val('View ' + data2[0].count + ' Comments');
    }
});

PHP:

---

在php端创建一个数组，用于保存答案1和答案2。然后，json_编码并回显新的多维数组，就像处理旧数组一样

然后在回调的javascript端运行JSON.parse（）将数组转换为javascript对象

例如：

if(isset($_GET['parcel_id'])) {
  $db = new ezSQL_mysql(DB_USER, DB_PASSWORD, DB_NAME, DB_HOST);
  $return = array();
  //get the apn based on id
  $data = $db->get_results("select apn from parcels where parcel_id=" . $_GET['parcel_id']);
  if($data != null){
    $return['data_one'] = $data;   
  }

  //count number of comments for the id
  $data2 = $db->get_results("select count(*) as count from comments where parcel_id=" .$_GET['parcel_id']);
  if($data2 != null){
    $return['data_two'] = $data2;
  }
  echo json_encode($return);
}

js:

---

我现在得到了“uncaughtsyntaxerror:unexpectedtokeno”，这可能是某种语法错误。可能是由于php错误，导致javascript试图读取的标记出错。因此，它没有得到xyz，而是读取了一条php错误消息，而jquery不知道如何解析它。

if(isset($_GET['parcel_id'])) {
  $db = new ezSQL_mysql(DB_USER, DB_PASSWORD, DB_NAME, DB_HOST);
  $return = array();
  //get the apn based on id
  $data = $db->get_results("select apn from parcels where parcel_id=" . $_GET['parcel_id']);
  if($data != null){
    $return['data_one'] = $data;   
  }

  //count number of comments for the id
  $data2 = $db->get_results("select count(*) as count from comments where parcel_id=" .$_GET['parcel_id']);
  if($data2 != null){
    $return['data_two'] = $data2;
  }
  echo json_encode($return);
}

success: function(res){
    //do stuff here on success
    res = JSON.parse(res);
    console.log(res);
}