Question

当[预订]中有数据时，我需要存储，但不存储任何内容以进行[扫描]

public function store(Request $request, $id){
    $event = event::findOrFail($id);
    $booking = booking::where('student_id',Auth::user()->student_id)
    ->where('name',Auth::user()->name)
    ->where('event_id',$event->id);
    if($booking!=NULL){
    $requestData = $request->all();
    scan::create($requestData);
    return redirect('event/' . $event->id .'/scan');
    return view('event.scan', compact('event','scan','booking'));
    }else{
    return redirect('event/' . $event->id .'/scan');
    }
}

但是使用$ booking == null时它存储了无法预订的任何东西

Answer 1

您的查询不完整。它缺少任何最终查询方法，例如get()，first()或exists()

在return块中，您还有两个if语句。只有第一个被执行，第二个将被忽略。

正确的代码应如下所示：

public function store(Request $request, $id){
    $event = event::findOrFail($id);

    $booking = booking::where('student_id',Auth::user()->student_id)
    ->where('name',Auth::user()->name)
    ->where('event_id',$event->id)
    ->get(); // look this line

    if($booking!=NULL){
        $requestData = $request->all();
        scan::create($requestData);
        return redirect('event/' . $event->id .'/scan'); // I think this should not be here, right?
        return view('event.scan', compact('event','scan','booking')); // this is being ignored.
    }else{
        return redirect('event/' . $event->id .'/scan');
    }
}

Answer 2

您可以像这样更新您的方法。它更清晰易读（无论如何，我认为）。

public function store(Request $request, $id)
{
    //use first to get a booking or use get() for collections of booking
    $booking = Booking::where('student_id', auth()->user()->student_id)
                ->where('event_id', $id)
                -first(); 

    if (is_null($booking)) {
        return redirect('event/' . $id . '/scan');
    }

    return view('event.scan', [
        'event' => Event::findOrFail($id),
        'scan' => Scan::create($request->all()),
        'booking' => $booking
    ]);
}

希望这会有所帮助干杯

无法有效地使用[IF]时无法存储

2 个答案: