Question

我有多个MySQL表，其中有许多行需要使用PHP在表中返回。我遇到的问题是当其中一个表有多个匹配ID时，如何正确显示信息。

以此为例。这是一个表，用于保存用户（userID）已预订的计划。

bk_schedule

id  userID  date    block   tos status
113 46  2013-12-31  3        yes    1
114 44  2013-12-26  1        yes    3
115 45  2013-12-31  1        yes    3
116 44  2013-12-31  2        yes    3
117 44  2013-12-31  1        yes    3

虽然它正在保存这些数据，但它还将数据保存到另一个表中，用户选择的“服务”分为他们选择的新行foreach服务。

bk_service

id  userID  bk_id   services
212 46       113    7
213 44       114    62
214 45       115    61
215 44       116    14
216 44       117    1
217 44       117    8
218 44       117    22
219 44       117    15

bk_id与bk_schedule id相关，以形成他们的关系。

现在，当我必须使用Laravel 4将此信息提取到表中时，如果我使用不同的表变量，我会将所有结果合并到每一行中。如果我尝试使用JOIN's使用相同的表集我得到的行很好但是它们循环遍历每个服务而不是组合（我猜它因为它循环每一行都将它计为一个新行）。

有点像这样。

userID  bk_id   services
44       116    14
44       114    62
44       117    8
44       117    22
44       117    15

以下是反映这一点的代码。

    public function showHistory($id) {

   $appointment = DB::table('bk_schedule')
    ->select('bk_schedule.id', 'bk_schedule.date', 'bk_timeslot.block', 'bk_status.status', 'pr_service.service')
    ->where('bk_schedule.userID', $id)      
    ->join('bk_status', 'bk_schedule.status', '=', 'bk_status.id')
    ->join('bk_timeslot', 'bk_schedule.block', '=', 'bk_timeslot.id')
    ->join('bk_service', 'bk_schedule.id', '=','bk_service.bk_id')
    ->join('pr_service', 'pr_service.id', '=', 'bk_service.services')
    ->orderBy('date', 'ASC')            
    ->get();

  // var_dump($appointment); die;

    $today = date('Y-m-d');

    foreach($appointment as $appointments) {

        $date = strtotime($appointments->date);     

        $appointments->date = date('l: F d, Y',$date);              
    }


    $service = DB::table('bk_service')
        ->select('pr_service.service', 'pr_service.price')
        ->join('pr_service', 'pr_service.id', '=', 'bk_service.services')
        ->where('bk_service.userID', $id)
        ->where('bk_service.bk_id', $appointments->id)
        ->get();

   return View::make('appointments.history', array('pageTitle' => 'Apppointment History',
                                'today' => $today, 'service' => $service,
                                'appointment' => $appointment)); 
}

刀片模板：

        <table class="main-table">
            <thead class="main-table-head">
                <th>Status/Result</th>
                <th>Date</th>
                <th>Block</th>
                <th>Services</th>
                <th>Action</th>
            </thead>
            <tbody class="main-table-head">
            @foreach($appointment as $appointments)
                <tr>
                    <td>{{{ $appointments->status }}}</td>
                    <td>{{{ $appointments->date }}}</td>
                    <td>{{{ $appointments->block }}}</td>
                    <td>
                        @foreach($service as $services)
                        {{{ $services->service }}}
                        @endforeach
                    </td>
                </tr>
            @endforeach
            </tbody>
        </table>

这基本上就是我想要的样子。（如果有帮助，这是约会历史页面）

userID  bk_id   services
44       117    1, 8, 22, 15
44       116    14
44       114    62

我试图尽可能详细地试图让它发挥作用是一种痛苦。我试过GROUP_CONCAT但是我遇到了同样的问题（它正在梳理该用户ID的所有记录）

我的尝试

        $schedule = DB::table('bk_schedule')
            ->select( DB::raw('users_information.street_2, users_information.phone_2, users_information.apartment, bk_schedule.note, bk_schedule.date, bk_schedule.office, bk_status.status, bk_schedule.id, bk_schedule.userID, bk_timeslot.block, users_information.last_name, users_information.street_1, users_information.phone_1, users_information.user_zip_code, group_concat(pr_service.short_name SEPARATOR " | ") as group_service, group_concat(pr_service.service SEPARATOR ", ") as service_detail'))
            ->join('users_information', 'bk_schedule.userID', '=', 'users_information.id')
            ->join('bk_timeslot', 'bk_schedule.block', '=', 'bk_timeslot.id')
            ->join('bk_service', 'bk_schedule.userID', '=', 'bk_service.userID')
            ->join('pr_service', 'bk_service.services', '=', 'pr_service.id')
            ->join('bk_status', 'bk_schedule.status', '=', 'bk_status.id')
            ->orderBy('bk_schedule.date', 'asc')
            ->groupBy('bk_schedule.id')
            ->paginate(15);

如果有人对我的最终解决方案感到好奇。

        $schedule = DB::table('bk_schedule')
            ->select( DB::raw('bk_schedule.office, pr_service.short_name, bk_timeslot.block, bk_schedule.date, bk_status.status, users_information.last_name, users_information.street_1, users_information.phone_1, users_information.user_zip_code, users_information.street_2, users_information.phone_2, users_information.apartment, bk_schedule.userID, bk_service.id, group_concat(pr_service.service)as service_detail, group_concat(pr_service.short_name)as group_service '))
            ->join('bk_service', 'bk_schedule.id', '=', 'bk_service.bk_id')
            ->join('users_information', 'bk_schedule.userID', '=', 'users_information.id')
            ->join('bk_status', 'bk_schedule.status', '=', 'bk_status.id')
            ->join('bk_timeslot', 'bk_schedule.block', '=', 'bk_timeslot.id')
            ->join('pr_service', 'bk_service.services', '=', 'pr_service.id')
            ->groupBy('bk_service.userID', 'bk_service.bk_id')
            ->paginate(15);

Answer 1

您需要按userId和预订ID进行分组。

select sc.userId, sc.id, group_concat(services)
  from bk_schedule sc
  join bk_service se on (sc.id = se.bk_id)
 group by sc.userId, sc.id;

在sqlfiddle

上查看

如何使用MySQL和PHP匹配两组数据

1 个答案: