Question

我有以下表格：

tbl_users
===============
uid
username
password
gid

tbl_groups
===============
gid
name
type

我试图弄清楚如何在f3中使用sqlmapper来查询其中username等于$ _POST [“username”]的两个表，并且能够获得组名和类型。是否可以使用此框架与sqlmapper一起加入类似的查询？

我一直在四处寻找，但却找不到任何例子。

Answer 1

您可以尝试为此设置一些虚拟字段：

$mapper->group_name = 'select name from tbl_groups where tbl_groups.gid=tbl_users.gid';
$mapper->group_type = 'select type from tbl_groups where tbl_groups.gid=tbl_users.gid';
$mapper->load(array('uid = ?',123));

echo $mapper->group_name;

Answer 2

以下是{VIEW}的示例：

我已经使用View view_user_list_with_referral

在这里实现了对SQL Mapper的分页

    $dropInstantWinnerView = $this->db->exec("DROP VIEW IF EXISTS view_user_list_with_referral;");
    $createInstantWinnerView = $this->db->exec("CREATE VIEW view_user_list_with_referral AS SELECT u.fb_id, fb_link, name, r.referred_by, u.created FROM users u LEFT OUTER JOIN referral r ON u.fb_id=r.joinee ");
    $user = new \DB\SQL\Mapper($this->db,'view_user_list_with_referral');
    $limit = 20;
    $page = Pagination::findCurrentPage();
    $order_condition = F3::get("PARAMS.order_condition");
    $order_class= "";
    if(!empty($order_condition)){
        $cond = explode(":", $order_condition);
        $option = array('order' => $cond[0].' '.$cond[1]);
        if($cond[1]=='ASC'){
            $order_condition = $cond[0].':DESC';
            $order_class = ":DESC";
        }else{
            $order_condition = $cond[0].':ASC';
            $order_class = ":ASC";
        }
    }else{
        $option = array('order' => 'created DESC');
    }

    $subset = $user->paginate($page-1, $limit, null, $option);

    $pages = new Pagination($subset['total'], $limit);
    $pages->setTemplate("admin/pagebrowser.html");

    F3::set('pagebrowser', $pages->serve());
    //echo "<pre>";print_r($subset);exit;
    F3::set('page', $page);
    F3::set('order_condition', $order_condition);
    F3::set('total_found_records', $user->count());

我希望它可以节省时间：）

f3框架sql查询加入

2 个答案:

以下是{VIEW}的示例：