Php 使用Flexigrid链接2个表格

嘿，伙计们，我遇到了点问题。我有两个表，我想链接以获取查找id的名称，这完全是直截了当的，但使用FlexGrid这样做是一件痛苦的事情，因为您应该以何种方式构建您的查询。这是我目前使用的代码，但是您得到了1000条记录的输出，这些记录应该只有3条

    //Setup sort and search SQL using posted data.
$sortSql = "ORDER BY $sortname $sortorder";
//$searchSql = ($qtype != '' && $query != '') ? "WHERE $qtype = '$query'" : 'WHERE tbl_p2e_place.CityID = tbl_lu_city.CityID';
$searchSql = ($qtype != '' && $query != '') ? "WHERE $qtype = '$query'" : '';

//Get total count of records
$sql = "SELECT COUNT(*) FROM tbl_p2e_place $searchSql";
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
$total = $row[0];

//Setup paging SQL
$pageStart = ($page -1)* $rp;
$limitSql = "LIMIT $pageStart, $rp";


//return JSON data
$data = array();
$data['page'] = $page;
$data['total'] = $total;
$data['rows'] = array();
$sql = "SELECT listingID, listingContractNumber, listingName, CityName 
        FROM tbl_p2e_place, tbl_lu_city
        $searchSql
        $sortSql
        $limitSql";
$results = mysql_query($sql);
while($row = mysql_fetch_assoc($results))
{
    $data['rows'][] = array(
    'id' => $row['listingID'],
    'cell' => array($row['listingID'], $row['listingContractNumber'],$row['listingName'], $row['CityName'])
    );
}

echo json_encode($data);

当我把

放在tbl_p2e_place.CityID=tbl_lu_city.CityID

中最符合逻辑的地方时，它仍然会这样做。我还有别的方法可以做到这一点吗？即使我必须使用jqgrid或任何同等技术

问候