我有以下ajax代码:
public function postInquire(Request $request)
{
$data = array(
'tripName' => $request->tripName,
'fullName' => $request->fullName,
'email' => $request->email,
'telephone' => $request->telephone,
'bodyMessage' => $request->message
);
Mail::to('mail@domain.com')->send(new TourInquire($data));
return response()
->json(['code'=>200,'success' => 'Your inquire is successfully sent.']);
}
我在控制器中的代码:
Route::post('/inquire', 'PostPublicController@postInquire')->name('postInquire');
我发布我的ajax表单的路线是:
<div class="modal-body">
<div id="form-messages" class="alert success" role="alert" style="display: none;"></div>
<div class="preview-wrap">
<img src="{{asset($tour->featuredImage->path)}}" class="preview-img thumbnail" alt="{{$tour->featuredImage->name}}">
<div class="form-wrap">
<form action="{{route('postInquire')}}" id="'#ajax-inquire" method="POST">
{{csrf_field()}}
<div class="form-group">
<input type="hidden" name="tripName" id="tripName" value="{{$tour->title}}">
<label>Name</label>
<input type="text" class="form-control" placeholder="Enter Your Full Name" name="fullName" id="fullName" required>
</div>
<div class="form-group">
<label>Email</label>
<input type="email" class="form-control" placeholder="Email Address" name="email" id="email" required>
</div>
<div class="form-group">
<label for="telephone">Phone</label>
<input type="tel" class="form-control" placeholder="Phone Number" name="telephone" id="telephone" required>
</div>
<div class="form-group">
<label for="message">Message</label>
<div class="row">
<textarea name="message" id="message" cols="30" rows="5" class="form-control"></textarea>
</div>
</div>
<button class="btn btn-primary hvr-sweep-to-right">Send Message</button>
</form>
</div>
</div>
</div>
使用上述代码,我可以通过ajax请求发送邮件。我想在我的视图中在警告框中显示json响应消息。但是我无法这样做,因为json响应消息显示在白页中,并且我的帖子路径为url。
视图中的HTML代码:
{{1}}
答案 0 :(得分:1)
HTML中有一个简单的拼写错误:
'
单引号#
和action
都不应该存在,并且表示您的jQuery选择器与表单不匹配,因此您的Javascript实际上都没有触发。表单只是正常提交,因此您最终会使用id
形式指定的网址。
id="ajax-inquire"
应指定为:
$()
附注:技术上并不重要,但您不需要在现有的jQuery对象上使用// Here you set form as a jQuery object
var form = $('#ajax-inquire');
// You don't need "$(form)" here, "form" is all you need
$(form).submit(function(event) {
// You can simply use the existing jQuery object
form.submit(function(event) {
。它的工作原理是因为jQuery接受一个jQuery对象作为选择器,但如果你没有以任何方式过滤或编辑选择器,它就是多余的。
select job_ID from JobCard Inner join Jobs On JobCard.job_ID = Job.job_ID where status = 'jobDone' group by Job.labor
答案 1 :(得分:0)
尝试将此返回:
$your_json= json_encode(['code'=>200,'success' => 'Your inquire is successfully sent.']);
return view('your_view',compact('your_json'));
然后在你的视图中:
{{json_decode($your_json)['success']}}