Laravel 验证器每次都失败,即使我已发布数据

Laravel Validator fails everytime even i has posted data

我是 laravel 的新手,正在尝试使用验证器 class。每次通过形成 post 数据都会失败。我用谷歌搜索和挖掘了很多东西以从互联网上获得答案,但 none 对我有用。请堆栈人员帮助我解决这个问题。

这是我的控制器

<?php

namespace App\Http\Controllers;

use Illuminate\Http\Request;
use Validator;

class AuthenticationController extends Controller
{
    //
    public function login()
    {
        return view('login');
    }
    public function register()
    {
        return view('register');
    }
    public function addUser(Request $request)
    {
        $Validator=Validator::make($request->all(),array('fname'=>'required'));
        if($Validator->fails())
        {
            return redirect()->back()->withInput()->withErrors($Validator->messages());
        }
    }
    
}

这是寄存器blade查看文件

<!DOCTYPE html>
<html>
<head>
  <link href="//maxcdn.bootstrapcdn.com/bootstrap/4.0.0/css/bootstrap.min.css" rel="stylesheet" id="bootstrap-css">
<script src="//maxcdn.bootstrapcdn.com/bootstrap/4.0.0/js/bootstrap.min.js"></script>
<script src="//cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>

<!------ Include the above in your HEAD tag ---------->
  <title>Login</title>
</head>
<body>


<div class="wrapper fadeInDown">
  <div id="formContent">
    <!-- Tabs Titles -->
                   
  <!-- Login Form -->
    <form method="POST" action="{{url('/register.submit')}}">
      <input type="text" id="first-name" class="fadeIn second" name="
      fname" placeholder="First Name">
      @if($errors->has('fname'))
      <div class="alert-danger">
      {{$errors->first('fname')}}
      </div>
      @endif
      @csrf
      
      <input type="text" id="last-name" class="fadeIn second" name="
      lastname" placeholder="Last Name">
      @if($errors->has('last_name'))
      <div class="alert-danger">
      {{$errors->first('last_name')}}
      </div>
      @endif

      <input type="text" id="email" class="fadeIn second" name="
      email" placeholder="Email">
      @if($errors->has('email'))
      <div class="alert-danger">
      {{$errors->first('email')}}
      </div>
      @endif
      
      <input type="password" id="password" class="fadeIn third" name="login" placeholder="password">
      @if($errors->has('password'))
      <div class="alert-danger">
      {{$errors->first('password')}}
      </div>
      @endif

      <input type="password" id="confirm-password" class="fadeIn third" name="confirmpassword" placeholder="Confirm Password">
      @if($errors->has('confirm_password'))
      <div class="alert-danger">
      {{$errors->first('confirm_password')}}
      </div>
      @endif

      <input type="submit" class="fadeIn fourth" value="Register">
    </form>

    <!-- Remind Passowrd -->
    <div id="formFooter">
      <a class="underlineHover" href="{{url('login')}}">Already have account SignIn</a>&nbsp;&nbsp;&nbsp;
      <a class="underlineHover" href="{{url('forget_password')}}">Forget Password</a>
    </div>

  </div>
</div>
</body>
</html>

确保您的 name 属性实际上在 一个 行中,并且不要分布在两行中。在此处显示代码标签中的内容将导致: name=" fname"

当您在 HTML 中显示此类字符串时,您很可能会看到空格被删除,但 PHP/Laravel 将其识别为空格,因此您的验证失败。

实际上进入视图文件编辑器 sublime 有类似

的换行代码
<input type="text" id="first-name" class="fadeIn second" name="
      fname" placeholder="First Name">

所以(我希望如此)laravel当我像

这样编辑代码时,把它当作name='fname'
<input type="text" id="first-name" class="fadeIn second" 
    name="fname" placeholder="First Name">

      

修改后效果很好。