Question

我目前有这段代码：

        return Datatable::query($query = DB::table('acquisitions')
            ->where('acquisitions.deleted_at', '=', null)
            ->where('acquisitions.status', '!=', 2)
            ->join('contacts', 'acquisitions.contact_id', '=', 'contacts.id')
            ->join('user', 'acquisitions.user_id', '=', 'user.id')
            ->select('contacts.*', 'acquisitions.*', 'acquisitions.id as acquisitions_id', 'user.first_name as supervisor_first_name', 'user.last_name as supervisor_last_name', 'user.id as user_id'))

来自用户表的数据用于2列：acquisitions.supervisor_id和acquisitions.user_id。我需要这两个表的first_name和last_name，但上述查询目前只使用id字段中的acquisitions.user_id。我也尝试使用表别名，这也行不通，我假设我在这里做错了。

简而言之：我还需要查询根据acquisitions.supervisor_id中的ID为用户选择数据，并将其设为supervisor_first_name和supervisor_last_name。

Answer 1

根据您对其他答案的最后评论，您需要每个参考表自我加入。试试这个：

$result = DB::select('SELECT
u.name user_first_name,
u.last_name user_last_name,
u.email user_email,
s.name supervisor_name,
s.last_name supervisor_last_name,
s.email supervisor_email
FROM acquisitions a
JOIN users u ON a.user_id = u.id
JOIN users s ON a.supervisor_id = s.id');

return $result;

请注意$result是StdClass个对象的数组，而不是Collection，但您仍然可以迭代它并调用当前项的值：

foreach ($result as $item) {
    print($item->supervisor_first_name);
}

如果您需要WHERE条款，例如要从acquisitions获取特定用户的行，您可以通过向查询添加参数来执行此操作，如下所示：

$result = DB::select('SELECT
u.name user_first_name,
u.last_name user_last_name,
u.email user_email,
s.name supervisor_name,
s.last_name supervisor_last_name,
s.email supervisor_email
FROM acquisitions a
JOIN users u ON a.user_id = u.id
JOIN users s ON a.supervisor_id = s.id
WHERE a.user_id = ?
', [3]);

修改

如果您需要将结果集设为Collection，则可以使用hydrate方法轻松将数组转换为1：

$userdata = \App\User::hydrate($result); // $userdata is now a collection of models

Answer 2

应该是这样的;

return Datatable::query($query = DB::table('acquisitions')
    ->where('acquisitions.deleted_at', '=', null)
    ->where('acquisitions.status', '!=', 2)
    ->join('contacts', 'acquisitions.contact_id', '=', 'contacts.id')
    ->join('user', 'acquisitions.user_id', '=', 'user.id')
            ->select( \DB::raw("  contacts.*, acquisitions.*, acquisitions.id as acquisitions_id, user.first_name as supervisor_first_name, user.last_name as supervisor_last_name, user.id as user_id   ") )
);

Answer 3

$devices = DB::table('devices as d')
->leftJoin('users as au', 'd.assigned_user_id', '=', 'au.id')
->leftJoin('users as cu', 'd.completed_by_user_id', '=', 'cu.id')
->select('d.id','au.name as assigned_user_name','cu.name as completed_by_user_name');

也请点击此链接

https://github.com/yajra/laravel-datatables/issues/161

Laravel将同一个表连接成2列

3 个答案: