2017-03-02 66 views
0

再回來 我有問題顯示列表數據從laravel 5.4(AdminLTE表單)我的註冊表中的數據庫。 有人幫我嗎?如何從註冊表格Laravel 5.4中的dropdownlist數據庫獲取數據?

編碼register.blade.php:

<div class="form-group"> 
        <label for="title">Pilih Kecamatan:</label> 
        <select name="kecamatan" id="kecamatan" class="form-control"> 
         <option value="">--- Pilih Kecamatan ---</option> 

        </select> 
       </div> 
       <div class="form-group"> 
        <label for="title">Pilih Desa:</label> 
        <select name="desa" id="desa" class="form-control"></select> 
       </div>  

編碼RegisterController.php

use RegistersUsers; 


public function showRegistrationForm() 
{ 
    return view('adminlte::auth.register'); 
} 


protected $redirectTo = '/home'; 


public function __construct() 
{ 
    $this->middleware('guest'); 
} 


protected function validator(array $data) 
{ 
    return Validator::make($data, [ 
     'name'  => 'required|max:255', 
     'username' => 'sometimes|required|max:255|unique:users', 
     'email' => 'required|email|max:255|unique:users', 
     'password' => 'required|min:6|confirmed', 
     'phone' => 'required', 
     'alamat' => 'required', 
     'kecamatan'=> 'required', 
     'desa'  => 'required', 
     'terms' => 'required', 
    ]); 
} 


protected function create(array $data) 
{ 
    $fields = [ 
     'name'  => $data['name'], 
     'email' => $data['email'], 
     'password' => bcrypt($data['password']), 
     'phone' => $data['phone'], 
     'alamat' => $data['alamat'], 

    ]; 
    if (config('auth.providers.users.field','email') === 'username' && isset($data['username'])) { 
     $fields['username'] = $data['username']; 
    } 
    return User::create($fields); 
} 

我需要一步如何獲取數據以及如何顯示它在我的寄存器域。 謝謝你們。

回答

0

控制器功能:

public function editProfile($id) 
{ 
    $surnames = DB::table('families')->lists('surname'); 
    return view('editprofile', ['surnames' => $surnames]); 
} 

查看:

<label for="lastname">Last Name</label> 
     <select class="form-control" name="lastname" id="lastname" data-parsley-required="true"> 
      @foreach ($surnames as $sn) 
      { 
      <option value="{{ $sn->surname }}">{{ $sn->surname }}</option> 
      } 
      @endforeach 
     </select> 

你應該通過https://laracasts.com/discuss/channels/laravel/fetch-dropdown-list-from-database-in-l-52?page=1

+0

我不能跑,也許是因爲輸精管版本laravel 方案,我使用laravel 5.4版本。 –

+0

你試過我剛給的方法嗎? –

+0

我已經做過之前,但是當我運行程序錯誤代碼是 姓氏未定義的變量。 –

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