我尝试使用POST方法发送表单,但我收到了查询异常错误。
SQLSTATE [23000]:完整性约束违规:1048列' post_id' 不能为空(SQL:插入
comments
(comment
,approved
,user_id
,post_id
,updated_at
,created_at
)值(sdfsdfsdff, 1,1,2017-12-04 07:18:13,2017-12-04 07:18:13))
' sdfsdfsdff'是我试图发送给"评论系统的评论",1是用户ID,空格是post id应该是的位置。
控制器
<?php
$this->validate($request, [
'comment' => 'required'
]);
$post = Post::find($post_id);
$comment = new Comment();
$comment->comment = $request->comment;
$comment->approved = true;
$comment->user_id = auth()->id();
$comment->post_id = $request->id;
$comment->save();
return response()->json([
'comment' => $comment->comment,
'user_id' => $comment->user_id,
'post_id' => $comment->post_id,
'post_slug' => $post->slug,
'success' => true
]);
视图中的Ajax代码
var urlPostComment = '{{ url('comments/store') }}';
$('.send').click(function(e){
e.preventDefault();
var dataSerialize = $('#comment-new').serialize();
$.ajax({
method: 'post',
url: urlPostComment,
data: dataSerialize,
dataType: 'json', // <- La coma
success: function (data) {
console.log(data); // <- Revisar JSON en la consola del navegador
if( data.success )
{
$('#comment-post').append(
'<p class="store-comment">' + data.comentario + '</p>'
);
}
}, // <- La coma
error: function () {
console.log('Error en el AJAX');
}, // <- La coma
complete: function () {
console.log('Termino el ajax');
}
});
});
表格
<div class="row">
<div class="col-xs-8 col-md-offset-2 form-create">
{!! Form::open(['route' => ['comments.store', $post->id], 'method' => 'POST', 'id' => 'add-comment']) !!}
<p class="error text-center alert alert-danger hidden"></p>
{{ Form::label('comment', ' Comentario', ['class' => 'glyphicon glyphicon-comment pull-right label-space', 'id' => 'comment-create']) }}
{{ Form::textarea('comment', null, ['class' => 'form-control comment-text', 'id' => 'comment-new', 'placeholder' => 'Escribe tu comentario']) }}
<p class="error text-center alert alert-danger hidden"></p>
{{ Form::submit('Agregar Comentario', ['class' => 'comment-message-guest-button send']) }}
{!! Form::close() !!}
</div>
</div>
评论HTML
<div class="row">
@foreach ($post->comments as $comment)
<section class="comment-list">
<article class="row">
<div class="col-md-3 col-sm-3 hidden-xs">
<figure class="thumbnail">
<img class="img-responsive" src="/uploads/avatars/{{ $comment->user->profilepic }}"/>
<figcaption class="text-center">{{ $comment->user->name }}</figcaption>
</figure>
</div>
<div class="col-md-8 col-sm-8">
<div class="panel panel-default arrow left">
<div class="panel-body">
<header class="text-left">
<div class="comment-user"><i class="fa fa-user"></i> {{ $comment->user->name }}
</div>
<time class="comment-date" datetime="{{ $comment->created_at->diffForHumans() }}"><i
class="fa fa-clock-o"></i> {{ $comment->created_at->diffForHumans() }}
</time>
</header>
<div id="comment-post" data-commentid="{{ $comment->id }}">
<p id="display-comment"
class="store-comment" {{ $comment->id }} {{ $post->id }}>{{ $comment->comment }}</p>
</div>
</div>
<div class="panel-footer list-inline comment-footer">
@if(Auth::guest())
No puedes responder ningún comentario si no has ingresado.
@else
@if(Auth::user() == $comment->user)
<a href="#" data-toggle="modal" data-target="edit-comment" class="edit-comment">Editar</a>
<a href="#" data-toggle="modal" data-target="delete-comment"
class="delete-comment">Eliminar</a>
@endif
@if(Auth::user() != $comment->user)
<a href="#">Responder</a>
@endif
@endif
</div>
</div>
</div>
</article>
</section>
@endforeach
</div>
主要错误:
POST http://localhost:8000/comments/store 500(内部服务器错误) &#34;来自Google Chrome控制台&#34;
在我的数据库评论表中,当我没有使用Ajax时,我确实有一个完美的外键,所以我不认为这是一个关系问题。提前谢谢。
答案 0 :(得分:1)
您的表单ID是“添加评论”,但在您的ajax中,您通过“comment-new”序列化您的表单。 并且您没有在表单中定义name =“id”字段,而是在控制器中。
$comment->post_id = $request->id;
答案 1 :(得分:0)
是的,你是对的。
这不是关系问题。但我觉得我有黑客攻击,
由于你在这里有$ post_id $post = Post::find($post_id);
,为什么你在这行使用$ request-&gt; id? $comment->post_id = $request->id;
我怀疑$ request-&gt; id没有帖子ID。
尝试使用这种方式:
$comment->post_id = $post_id;
答案 2 :(得分:0)
我认为你的路线看起来像这样
Route::post('comments.store/{id}', ...);
您可以通过
访问请求中的路由参数{id}
$request->route('id');
但由于你在控制器内部,你可以只访问store
函数中的参数
public function store($id) {
echo $id;
}