Ajax自动完成搜索错误

我试图让我的自动完成搜索工作，但当我试图从json数组中获取数据时，它不起作用。我对此不太熟悉，但我的代码就是这样的。我想我知道第一个文件是如何工作的，但我不知道我的"searchapi.php"将如何编写。这张$.地图是什么？如果有人能解释一下就太好了：D谢谢

index.php:
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script src="http://code.jquery.com/ui/1.10.3/jquery-ui.js"></script>
<link rel="stylesheet" href="/resources/demos/style.css" />
   <style>
     .ui-autocomplete-loading {
        background: white url('images/ui-anim_basic_16x16.gif') right center no-repeat;
      }
      #city { width: 15em; }
      </style>  
    <script>
          $(function() {
            function log( message ) {
              $( "<div>" ).text( message ).prependTo( "#log" );
              $( "#log" ).scrollTop( 0 );
            }
            $( "#city" ).autocomplete({
              source: function( request, response ) {
                $.ajax({
                  url: 'searchapi.php',
                  dataType: "jsonp",
                  data: {
                    featureClass: "P",
                    style: "full",
                    maxRows: 12,
                    name_startsWith: request.term
                  },
                  success: function( data ) {
                    response( $.map(data.table(?), function( item ) {
                      return {
                        label: item.name,
                        value: item.name
                      }
                    }));
                  }
                });
              },
              minLength: 2,
              select: function( event, ui ) {
                log( ui.item ?
                  "Selected: " + ui.item.label :
                  "Nothing selected, input was " + this.value);
              },
              open: function() {
                $( this ).removeClass( "ui-corner-all" ).addClass( "ui-corner-top" );
              },
              close: function() {
                $( this ).removeClass( "ui-corner-top" ).addClass( "ui-corner-all" );
              }
            });
          });
          </script>

searchapi.php

  <?php 
  $host = "localhost";
  $user = "root";
  $pass = "";
  $databaseName = "mydb";

$con = mysql_connect($host,$user,$pass);
$dbs = mysql_select_db($databaseName, $con);

$value = @$_POST['name_startsWith'];
$data = ("select name from column where name LIKE '".$value."'");
$result = mysql_query($data);
$dataArray = array();
while($array = mysql_fetch_assoc($result)){
    $dataArray[] = $array;
} 
echo json_encode($dataArray);

?>