如何在5.2中进行验证?

时间:2016-09-10 02:05:56

标签: laravel-5.2

任何人都可以对这种类型验证有所了解。

 public function joinclass()
    {
      if($class_code = $request->get('class_code');
        $classroom = classroomModel::where('class_code',$class_code)->first();
       {
          $class = new joinclass();
        $class -> user_id = Auth::user()->id;
        $class -> class_code = $request['class_code'];
        $class -> save();
       }else if{
         Flash::message('Your code is not found in databse');
        }elseif($classroom = joinclass::where('class_code',$class_code)->first();){
      Flash::message('You are already in this classroom');
}
    }

此代码和此类型条件的写入formet是什么。

1 个答案:

答案 0 :(得分:1)

您可以使用Validator类来验证请求 方法如下:

public function joinclass(Request $request)
{
  //Create a Validator for your request
  $validator = Validator::make($request->all(), [
        'class_code' => 'required',// use exists validation to validate if an entry exists in the table with value provided "exists:table,column"
        //other required validations
    ]);

  //validate the request
  if ($validator->fails()) {
  //if validation fails return the error 
        return Redirect::back()
            ->withErrors($validator);
  }
  else{
  //if validation passes
    $classroom = classroomModel::where('class_code',$class_code)->first();
    if($classroom = joinclass::where('class_code',$class_code)->first();){
        Flash::message('You are already in this classroom');
    }else{
        $class = new joinclass();
        $class -> user_id = Auth::user()->id;
        $class -> class_code = $request['class_code'];
        $class -> save();
    }       
  }   
}