Php 如果存在验证错误,如何打开模式?
我有以下代码,如果用户未经身份验证,则显示此“登录”链接: 模态:Php 如果存在验证错误,如何打开模式?,php,laravel,Php,Laravel,我有以下代码,如果用户未经身份验证,则显示此“登录”链接: 模态: <div class="modal fade" id="login_modal" tabindex="-1" role="dialog" aria-labelledby="myLargeModalLabel" aria-hidden="true"> <div class="modal-dialog modal-sm"> <div class="modal-content">
<div class="modal fade" id="login_modal" tabindex="-1" role="dialog" aria-labelledby="myLargeModalLabel" aria-hidden="true">
<div class="modal-dialog modal-sm">
<div class="modal-content">
<div class="modal-header">
<h5 class="modal-title" id="exampleModalLabel">Login</h5>
<button type="button" class="close">
<span aria-hidden="true">×</span>
</button>
</div>
<div class="modal-body">
<div class="container">
<div class="row">
@include('includes.errors')
<form class="clearfix" method="POST" action="{{ route('login') }}">
{{ csrf_field() }}
<div class="form-group col-12 px-0">
<label for="inputEmail4">Email</label>
<input type="email" class="form-control" value="{{ old('email') }}" name="email" required autofocus placeholder="Email">
</div>
<div class="form-group col-12 px-0">
<label for="inputEmail4">Password
<a href="{{ route('password.request') }}" class="text-gray ml-1" style="text-decoration: underline">
<small>Recover password</small></a> </label>
<input type="password" class="form-control" name="password" required placeholder="Palavra-passe">
</div>
<button type="submit" class="btn btn-primary btn d-block w-100">Login</button>
</form>
</div>
</div>
</div>
<div class="modal-footer">
<button type="button" id="close_login_modal" class="btn btn-primary" data-dismiss="modal">Close</button>
</div>
</div>
</div>
</div>
登录
&时代;
@include('includes.errors')
{{csrf_field()}}
电子邮件
密码
登录
接近
errors.blade.php
@if ($errors->any())
<ul>
@foreach ($errors->all() as $error)
<li>{{ $error }}</li>
@endforeach
</ul>
@endif
@if($errors->any())
@foreach($errors->all()作为$error)
- {{$error}}
@endforeach
@恩迪夫
您应该使用ajax,当您使用php发布时,它会刷新页面。谢谢,但是它不适用于类似“@if($errors->any())alert(“test”);$(“#login_modal”)。on('show.bs.modal',function(){})@endif”之类的东西?如果这不起作用,警报会显示,但模式不会打开。您也可以通过javascript通过yourResponse.responseJSON
获取错误。无论如何,如果你想做点击操作,你只需像这样触发点击即可。触发('click')在你的情况下,它将是$('show#login_modal')。触发('click')
我猜。谢谢,但也不适用于“@if($errors->any())警报('test”);$('show#login_modal')。触发('click');@endif“。但警报可以工作。
<div class="modal fade" id="login_modal" tabindex="-1" role="dialog" aria-labelledby="myLargeModalLabel" aria-hidden="true">
<div class="modal-dialog modal-sm">
<div class="modal-content">
<div class="modal-header">
<h5 class="modal-title" id="exampleModalLabel">Login</h5>
<button type="button" class="close">
<span aria-hidden="true">×</span>
</button>
</div>
<div class="modal-body">
<div class="container">
<div class="row">
@include('includes.errors')
<form class="clearfix" method="POST" action="{{ route('login') }}">
{{ csrf_field() }}
<div class="form-group col-12 px-0">
<label for="inputEmail4">Email</label>
<input type="email" class="form-control" value="{{ old('email') }}" name="email" required autofocus placeholder="Email">
</div>
<div class="form-group col-12 px-0">
<label for="inputEmail4">Password
<a href="{{ route('password.request') }}" class="text-gray ml-1" style="text-decoration: underline">
<small>Recover password</small></a> </label>
<input type="password" class="form-control" name="password" required placeholder="Palavra-passe">
</div>
<button type="submit" class="btn btn-primary btn d-block w-100">Login</button>
</form>
</div>
</div>
</div>
<div class="modal-footer">
<button type="button" id="close_login_modal" class="btn btn-primary" data-dismiss="modal">Close</button>
</div>
</div>
</div>
</div>
@if ($errors->any())
<ul>
@foreach ($errors->all() as $error)
<li>{{ $error }}</li>
@endforeach
</ul>
@endif