Php 更新laravel表单中按钮点击的数据库

Php 更新laravel表单中按钮点击的数据库,php,database,forms,laravel-4,Php,Database,Forms,Laravel 4,我有这个想法 {{ Form::open() }} @if($order->paid == 0) <button type="submit" class="btn btn-primary" value="1">Mark Order as Paid</a> @else <button type="submit" class="btn btn-primary" value="0">Mark Order as Unpaid</a>

我有这个想法

{{ Form::open() }}
@if($order->paid == 0)
    <button type="submit" class="btn btn-primary" value="1">Mark Order as Paid</a>
@else
    <button type="submit" class="btn btn-primary" value="0">Mark Order as Unpaid</a>
@endif
{{ Form::close() }}

是否可以这样给按钮
0
1
赋值,然后单击以更新数据库列?当前未更新,但如何进行更新?

您可以在每个按钮中放置一个隐藏字段,如下所示:

{{ Form::open() }}
@if($order->paid == 0)
    {{ Form::hidden('paid', 1, ['id' => 'paid']) }}
    <button type="submit" class="btn btn-primary" value="1">Mark Order as Paid</a>
@else
    {{ Form::hidden('paid', 0, ['id' => 'paid']) }}
    <button type="submit" class="btn btn-primary" value="0">Mark Order as Unpaid</a>
@endif
{{ Form::close() }}
惠特

你的表单按钮也能用

{{ Form::open() }}
@if($order->paid == 0)
    {{ Form::hidden('paid', 1, ['id' => 'paid']) }}
    <button type="submit" class="btn btn-primary" value="1">Mark Order as Paid</a>
@else
    {{ Form::hidden('paid', 0, ['id' => 'paid']) }}
    <button type="submit" class="btn btn-primary" value="0">Mark Order as Unpaid</a>
@endif
{{ Form::close() }}
<input id="paid" name="paid" type="hidden" value="1">
and
<input id="paid" name="paid" type="hidden" value="0">
$paid = Input::get('paid');
$order->paid = Input::get('paid');