我已经创建了一个使用Laravel5.5和Mailtrap.io联系形式,这是很好的工作。但是,我正在尝试使联系人表单发送,而不重新加载页面来完成此操作--我正在使用AJAX。但是,我不知道如何将AJAX与Laravel和Mailtrap.io连接起来。
这就是我创造的形式。
@if(Session::has('success'))
<p class="alert {{ Session::get('alert-class', 'success') }}">{{ Session::get('success') }}</p>
@endif
<form action="{{ url('contact') }}" method="POST" class="wpcf7-form" novalidate="novalidate">
{{ csrf_field() }}
<div class="primary-form">
<div class="col-xs-6">
<span class="wpcf7-form-control-wrap text-768"><input type="text" name="first-name" value="" size="40" class="wpcf7-form-control wpcf7-text wpcf7-validates-as-required" placeholder="First Name" /></span>
</div>
<div class="col-xs-6">
<span class="wpcf7-form-control-wrap text-768"><input type="text" name="last-name" value="" size="40" class="wpcf7-form-control wpcf7-text wpcf7-validates-as-required" aria-required="true" aria-invalid="false" placeholder="Last Name" /></span>
</div>
<div class="col-xs-6">
<span class="wpcf7-form-control-wrap email-766"><input type="email" name="email" value="" size="40" class="wpcf7-form-control wpcf7-text wpcf7-email wpcf7-validates-as-required wpcf7-validates-as-email" aria-required="true" aria-invalid="false" placeholder="E-mail Address" /></span>
</div>
<div class="col-xs-6">
<span class="wpcf7-form-control-wrap phone"><input type="text" name="phone" value="" size="40" class="wpcf7-form-control wpcf7-text wpcf7-validates-as-required" aria-required="true" aria-invalid="false" placeholder="Phone Number" /></span>
</div>
<div class="col-xs-12">
<span class="wpcf7-form-control-wrap textarea-6"><textarea name="message" cols="40" rows="10" class="wpcf7-form-control wpcf7-textarea" aria-invalid="false" placeholder="Your Message"></textarea></span>
</div>
<div class="pull-right" style="margin-right:10px;">
<p class="clearfix">
<button type="submit" class="button button-simple btn-submit mt-30">SEND</button>
</p>
</div>
</div>
</form>我使用的AJAX方法不需要重新加载页面就可以发送页面。
$(".btn-submit").click(function(e){
e.preventDefault();
var first_name = $("input[name=first-name]").val();
var last_name = $("input[name=last-name]").val();
var email = $("input[name=email]").val();
var phone = $("input[name=phone]").val();
var bodyMessage = $("textarea[name=message]").val();
$.ajax({
type:'POST',
url:'/contact',
data:{first_name:first_name, last_name:last_name, email:email, phone:phone, bodyMessage:bodyMessage},
success:function(data){
alert(data.success);
}
});
});这是我的ContactController。
<?php
namespace App\Http\Controllers;
use Illuminate\Http\Request;
use App\Post;
use Mail;
use Session;
class ContactController extends Controller
{
public function postContact(Request $request){
$this->validate($request, ['email' => 'required|email'] );
$data = array(
'first_name' => $request->first_name,
'last_name' => $request->last_name,
'email' => $request->email,
'phone' => $request->phone,
'bodyMessage' => $request->message
);
Mail::send('emails.contact', $data, function($message) use ($data){
$message->from($data['email']);
$message->to('jafar@calmcollective.co.uk');
$message->subject('Contact Details');
});
Session::flash('success', 'Your E-mail was sent! Allegedly.');
Session::flash('alert-class', 'alert-success');
return redirect('contact');
}
}我花了好几个小时在这上面。任何帮助都将不胜感激。
发布于 2018-02-19 16:40:49
从您对最初问题的回答中,我可以看到您的ContactController方法的拼写错误:
Route::post('/contact', 'ContactController@PostContact');应该是
Route::post('/contact', 'ContactController@postContact');然后,在javascript中,您应该侦听表单提交事件,而不仅仅是submit按钮单击的事件,并将csrf_token添加到数据中:
$("form.wpcf7-form").submit(function(e){
e.preventDefault();
var token = $("input[name=_token]").val(); // The CSRF token
var first_name = $("input[name=first-name]").val();
var last_name = $("input[name=last-name]").val();
var email = $("input[name=email]").val();
var phone = $("input[name=phone]").val();
var bodyMessage = $("textarea[name=message]").val();
$.ajax({
type:'POST',
url:'/contact',
dataType: 'json',
data:{_token: token, first_name:first_name, last_name:last_name, email:email, phone:phone, bodyMessage:bodyMessage},
success:function(data){
alert(data.success);
}
});
});最后,您应该在ContactController中返回一个成功参数:
<?php
namespace App\Http\Controllers;
use Illuminate\Http\Request;
use App\Post;
use Mail;
use Session;
class ContactController extends Controller
{
public function postContact(Request $request){
$this->validate($request, ['email' => 'required|email'] );
$data = array(
'first_name' => $request->first_name,
'last_name' => $request->last_name,
'email' => $request->email,
'phone' => $request->phone,
'bodyMessage' => $request->message
);
Mail::send('emails.contact', $data, function($message) use ($data){
$message->from($data['email']);
$message->to('jafar@calmcollective.co.uk');
$message->subject('Contact Details');
});
return response()->json(['success' => 'Your E-mail was sent! Allegedly.'], 200);
}
}发布于 2018-02-19 15:55:49
首先,表单正在重新加载,因为您在提交按钮上使用了单击处理程序,希望防止默认设置:$(".btn-submit").click(function(e),将单击更改为submit:$("#form-id").submit(function(e)。
其次,为什么不发送FormData对象而不是所有这些:
e.preventDefault();
var first_name = $("input[name=first-name]").val();
var last_name = $("input[name=last-name]").val();
var email = $("input[name=email]").val();
var phone = $("input[name=phone]").val();
var bodyMessage = $("textarea[name=message]").val();
$.ajax({
type:'POST',
url:'/contact',
data:{first_name:first_name, last_name:last_name, email:email, phone:phone, bodyMessage:bodyMessage},
success:function(data){
alert(data.success);
}
});试试这个:
e.preventDefault();
var form = document.querySelector('#form-id');
var data = new FormData(form);
$.ajax({
type:'POST',
url:'/contact',
data:data,
success:function(data){
alert(data.success);
}
});在php中,您必须通过表单字段的名称访问变量。
编辑
要明确使用请求对象访问php脚本中的变量,请执行以下操作:$request->input('nameOfInputfield'),要访问数组中的所有值,请使用$request->all()
发布于 2018-02-19 15:55:57
我经常用jquery和jquery表单插件来完成这个任务:
http://malsup.com/jquery/form/ jquery表单插件
https://stackoverflow.com/questions/48869109
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