无法将数组从php返回到ajax

我在php页面中有一个按钮，调用onclick ajax函数。然后我用php来更新我的数据库并将结果返回给ajax。问题是由于某种原因我无法返回结果。我总是得到错误函数，最糟糕的是我没有在我的警报框中得到'returned.msg'值只是'undefined`

I have a button in php page that calls onclick a ajax function. Then I use php to update my database and return the result back to ajax. The problem is that for some reason I cannot return the result. I always get the error function and worst I don't get the 'returned.msg' value just `undefined` in my alert box

function AjaxAddPerson() {
$.ajax({
        type: "GET",
        url: "../lib/addPerson.php",
        dataType: 'json',
        data: { person: document.getElementById("PersonSearchInput").value},
        success:function( returned ) {
         alert( "Data Saved: " + returned.msg );
        },
        error:function(returned ) {
            alert( "Could not Saved: " +  returned.msg );
        }
       });
}

<?php
	session_start();
    define("DB_HOST", '.000webhost.com');
    define("DB_USER", '_');
    define("DB_PASSWORD", '');
    define("DB_DATABSE", '_');

    $email = $_GET['person'];
    $UserEmail = $_SESSION['login'];

    $conn = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
    mysql_select_db(DB_DATABSE, $conn);

    $sql_find = mysql_query("SELECT * FROM Users WHERE UserEmail='".$email."'");

    $UserInfo = array();

    while ($row = mysql_fetch_array($sql_find, MYSQL_ASSOC)) {
      $UserInfo[] =  $row['UserEmail'];  
    }
    
    if(in_array($email, $UserInfo)) {
        $result = mysql_query("UPDATE `Users` SET `Contacts` = '".$email."' WHERE `UserEmail` = '".$UserEmail."'");
	    if ($result) {
           $returnRes = array( 'found' => true, 'msg' => "Person added to your record");
		   echo json_encode($returnRes);
		} 
		else {
           $returnRes = array( 'found' => false, 'msg' => "Error adding person to your record \n Is the person emails' correct?");
		   echo json_encode($returnRes);
		}
        
    }
    else {
        $returnRes = array( 'found' => false, 'msg' => "We couldn't find the user in our databases.");
    	echo json_encode($returnRes);
    }

?>

PS。

代码不是最好的，容易受到sql注入攻击。但这不是我现在的问题。

PS.
The code is not the best and is vulnerable to sql-injection. But this isn't my problem for now.