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嗨，这是我的控制器功能。在我看来，我将一个来自数据库的电子邮件变量设置为会话数据。我在控制器中验证了它是否正确。我得到它完美。但是，当我尝试将该电子邮件 ID 作为 TO 字段变量发送时，它不起作用。

function do_upload()
{ 
    $config['upload_path'] = 'C:\xampp\htdocs\upload\application';
    $config['allowed_types'] = 'doc|docx|pdf';
    $config['max_size'] = '10240';
    $email = $this->session->userdata("employer_email");// If i echo $email; I will get my passed session variable

    $this->load->library('upload', $config);

    if ( ! $this->upload->do_upload())
    {   
        $error = array('error' => $this->upload->display_errors());

        $this->load->view('upload_form', $error);
    }
    else
    { 
        $data = array('upload_data' => $this->upload->data()); 

        echo "Resume Successfully uploaded to database.............";
        $file = $data['upload_data']['full_path'];

    $config = Array(
      'protocol' => 'smtp',
      'smtp_host' => 'ssl://smtp.googlemail.com',
      'smtp_port' => 465,
      'smtp_user' => 'xxx@gmail.com', // change it to yours
      'smtp_pass' => 'xxxx', // change it to yours
      'mailtype' => 'html',
      'charset' => 'iso-8859-1',
      'wordwrap' => TRUE
    );


      $this->load->library('email', $config);
      $this->email->set_newline("\r\n");
      $this->email->from('xxx@gmail.com'); // change it to yours
      $this->email->to($email); // But here it's not working
      $this->email->subject('Email using Gmail.');
      $this->email->message('Working fine ! !');
      $this->email->attach($file);
      if($this->email->send())
     {
      echo 'Email sent.';
     }
     else
    {
     show_error($this->email->print_debugger());
    }
    }


}
4

1 回答 1

1

您的 $email 正在传递给没有引号的函数。想想它是如何从会话中产生的。如果会话中的电子邮件是 xxx@gmail.com，那么您将传递以下内容：

$this->email->to(xxx@gmail.com);

您希望该行显示为：

$this->email->to('xxx@gmail.com');

试试这个：

$email = '"'.$this->session->userdata("employer_email").'"';
于 2012-11-01T06:57:31.950 回答