Laravel：如何将值从表单传递给控制器？
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Laravel: how do I pass a value from a form to a controller?
提问by Josh
I have a form:
我有一个表格：
{ Form::open(array('action' => 'RatesController@postUserRate', $client->id)) }}
{{ Form::text('rate', '', array('placeholder' => 'Enter new custom client rate...')) }}
{{ Form::submit('Submit', array('class' => 'btn btn-primary')) }}
{{ Form::close() }}
How do I pass my $client->id value through the form to my controller method?
如何通过表单将我的 $client->id 值传递给我的控制器方法？
I currently have a controller method that looks like this:
我目前有一个控制器方法，如下所示：
public function postUserRate($id)
{
$currentUser = User::find(Sentry::getUser()->id);
$userRate = DB::table('users_rates')->where('user_id', $currentUser->id)->where('client_id', $id)->pluck('rate');
if(is_null($userRate))
{
...
}else{
....
}
}
And the error log says "Missing argument 1 for RatesController::postUserRate()"
错误日志显示“RatesController::postUserRate() 缺少参数 1”
Any ideas on how to pass this $client->id into my controller so I can use it as I want to above?
关于如何将这个 $client->id 传递到我的控制器以便我可以像上面那样使用它的任何想法？
回答by JeffreyWay
Add {{ Form::hidden('id', $client->id) }}to the form. Then, when it's posted, you can fetch its value per usual with Input::get('id').
添加{{ Form::hidden('id', $client->id) }}到表单中。然后，当它发布时，您可以使用Input::get('id').
Also, remove the postUserRatemethod's argument.
此外，删除postUserRate方法的参数。
回答by zianwar
You simply use :
您只需使用：
Form::open(array('action' => array('Controller@method', $user->id)))
Form::open(array('action' => array('Controller@method', $user->id)))
- the variable
$user->idis passed as argument to the methodmethod, also this last one should recieve an argument as well, like so :method($userId)
- 变量
$user->id作为参数传递给方法method，最后一个也应该接收一个参数，如下所示：method($userId)
Source : Laravel documentation
来源：Laravel文档
