如果郵件發送成功或錯誤Codeigniter AJAX如何返回

Question

我想要返回一個值從控制器返回到ajax發送電子郵件的時間和地點。以下是一些代碼。

在下面的代碼，我想設置在會話中的閃存數據，並應設置的任一條件，並呼籲它回到了AJAX

控制器

public function send_reply(){ 
     $data = array(
     'from_email' => "jeevanseq@gmail.com", 
     'recipient' => $this->input->post('recipient'), 
     'from_name' => "jeeva", 
     'subject' => $this->input->post('subject'), 
     'message' => $this->input->post('message'), 
     ); 

     echo json_encode($data); 
     $this->load->library('email'); 

     $this->email->from($data['from_email'], $data['from_name']); 
     $this->email->to($data['recipient']); 
     $this->email->subject($data['subject']); 
     $this->email->message($data['message']); 

     //Send mail 
     if($this->email->send()) { 
      $this->session->set_flashdata("email_sent","<span> * Email sent successfully. </span> "); 

     //$this->load->view('adminhp'); 
     } else { 

      $this->session->set_flashdata("email_sent","<span> * Error in sending Email.</span> "); 
     // $this->load->view('adminhp'); 


     } 
     } 

Ajax代碼

// Ajax post for email sending 
$(document).ready( function() { 
$(".mailsub").click(function(event) { 

event.preventDefault(); 
var recipient = jQuery(this).closest('form').find('input.recipient').val(); 
var subject = jQuery(this).closest('form').find('input.subject').val(); 
var message = jQuery(this).closest('form').find('textarea.message').val(); 

jQuery.ajax({ 
type: "POST", 
url: "<?php echo base_url(); ?>" + "index.php/user_authentication/send_reply", 
dataType: 'json', 
data: {recipient: recipient, subject: subject, message:message}, 
success: function(result) { 
if (result) 
{ 
// Show Entered Value 
jQuery("div#result").show(); 
jQuery("div#rslt").html("Sent Successfully"); 
console.log("Success"); 

} 
} 
}); 
}); 
}); 

我想將它設置爲fla在下面的代碼SH數據

echo "<br/>"; 
     echo "<div id='result' style='display: none'>"; 
     echo "<div id='rslt' style='display:inline'></div>"; 
     // echo $this->session->flashdata('email_sent'); 
     echo "</div>"; 
     echo "<br/>"; 

Answer 1

首先在控制器，你必須呼應消息阿賈克斯

$msg = ''; 
if($this->email->send()) { 
    $msg = 'Email sent successfully'; 
} else { 
    $msg ='Error in sending'; 
    } 
echo $msg;die; 

第二隻輸入反應添加到相應的div：

success: function(result) { 
      if (result) 
      { 
       console.log(result); 
       jQuery("div#result").show(); 
       jQuery("div#rslt").html(result); 

      }