如何laravel 5.4.24验证蛞蝓(How to validation slug in lar

如何在laravel创造独特的蛞蝓和验证它们？

这里是我的验证代码：

$this->validate($request,[
        'company_name' => 'required|unique:admin_users,company_name,slug|max:191',
    ]);

这里是我的蛞蝓代码：

$db_filed->company_name = str_slug($request->company_name, '-');

谢谢。

Answer 1:

设置一个FormRequest做的，像这样的规则路由的验证：

https://laravel.com/docs/5.4/validation#form-request-validation

public function rules()
{
    return [
         'company_name' => 'required|unique:admin_users,company_name,slug|max:191'
    ];
}

或者你需要在分配给该公司的名称之前创建蛞蝓。

https://laravel.com/docs/5.4/validation#manually-creating-validators

$slug = str_slug($request->company_name, '-');

$validator = Validator::make(['company_name' => $slug], [
    'company_name' => 'required|unique:admin_users,company_name,slug|max:191'
]);

if (!$validator->fails()) {
    $db_filed->company_name = $slug;
    $db_filled->save();
}

Answer 2:

我想这样的，现在它的工作，

下面是代码形式：

 <div class="form-group">
    <input type="text" class="form-control" placeholder="Company Name" name="company_name" value="{{ ucwords(str_replace('-',' ',old('company_name'))) }}" required>
 </div>

下面是控制器代码：

public function store(Request $request)
{
    $request['company_name'] = str_slug($request->company_name, '-');

    $this->validate($request,[
        'company_name' => "required|unique:admin_users,company_name|max:191",
    ]);
    $db_filed = new AdminUser;
    $db_filed->company_name = $request->company_name; 

    $db_filed->save();
}

Answer 3:

你可以创建你的控制器内塞可能是存储功能像这里面的

public function store(CompanyNameRequest $request)
{
    $slug = uniqid();
    $ticket = new CompaanyName(array(
        'title' => $request->get('title'),
        'content' => $request->get('content'),
        'slug' => $slug
    ));

    $ticket->save();

    return redirect('/contact')->with('status', 'Your order is been proccess! Its unique id is: '.$slug);

}

文章来源: How to validation slug in laravel 5.4.24