Question

我的控制器功能为：

public function verify()
    {
    $token= $this->uri->segment(3);
    $email_verification=$this->load->site_model->verifyEmailAddress($token);

        if ($email_verification=== FALSE)
        {
            redirect('site/index');
        }
        else
        {



$type=$this->load->site_model->select1($token);
            //print_r($type);
            if($type['user_type']=="Employer"){
                redirect('site/emp');
                }
                else{
                    redirect('site/seek');
                }

        }
    }

我的模特是：

<?php
class Site_model extends CI_Model
{
    public function __construct()
    {
        parent:: __construct();
        $this->load->database();

    }
    public function insert($token)
    { 
        $data = array(
        'name'=>$this->input->post('name'),
        'email'=>$this->input->post('email'),
        'phone'=>$this->input->post('phone'),
        'user_type'=>$this->input->post('utype'),
        'token'=>$token,
        );

        $this->db->insert('tbl_user',$data);

        $email=$this->input->post('email');
        $name=$this->input->post('name');
        $html ="http://localhost/jobs/hmvc/index.php/site/verify/".$token;
        $config = array(
       'protocol'  => 'smtp',
       'smtp_host' => 'ssl://smtp.googlemail.com',
       'smtp_port' => '465',
       'smtp_user' => 'someone@gmail.com',
       'smtp_pass' => 'something',
       'mailtype'  => 'html',
       'starttls'  => true,
       'newline'   => "\r\n"
       );
        $this->load->library('email',$config);
        $this->email->From("someone@gmail.com");
        $this->email->to($email);
        $this->email->subject('test');
        $this->email->message('<b>Hi '.$name.' </b><p>Welcome! You’re almost done.!Click the link to confirm your email address..</p>'.$html);
        //$this->email->send();

        if($this->email->send()) {
            echo '<script>alert("Email sent successfully")</script>';
            } else {
        $this->email->print_debugger();
        }

    }


    public function verifyEmailAddress($token)  
        {  
        $data=array('email_verification'=>1);
               $this->db->where('token',$token); 
               $this->db->update('tbl_user',$data);
                  return true;

        }
public function select1($token) {         
                  $this->db->select('user_type');
                  $this->db->from('tbl_user');
                  $this->db->where('token',$token);
                  return $this->db->get()->row();
    }


?>

当运行代码的em显示为致命错误的错误：不能在第89行的C：\ xampp \ htdocs \ jobs \ hmvc \ application \ modules \ site \ controllers \ site.php中使用stdClass类型的对象< / p>
我该如何解决这个问题？ codeigniter中的这个错误是什么意思？

Answer 1

 <?php  
 public function insert($token)
    { 
        $data = array(
        'name'=>$this->input->post('name'),
        'email'=>$this->input->post('email'),
        'phone'=>$this->input->post('phone'),
        'user_type'=>$this->input->post('utype'),
        'token'=>$token,
        );

        $this->db->insert('tbl_user',$data);
        $ret =$this->db->insert_id();


        $email=$this->input->post('email');
        $name=$this->input->post('name');
        $html ="http://localhost/jobs/hmvc/index.php/site/verify/".$token;
        $config = array(
       'protocol'  => 'smtp',
       'smtp_host' => 'ssl://smtp.googlemail.com',
       'smtp_port' => '465',
       'smtp_user' => 'someone@gmail.com',
       'smtp_pass' => 'something',
       'mailtype'  => 'html',
       'starttls'  => true,
       'newline'   => "\r\n"
       );
        $this->load->library('email',$config);
        $this->email->From("someone@gmail.com");
        $this->email->to($email);
        $this->email->subject('test');
        $this->email->message('<b>Hi '.$name.' </b><p>Welcome! You’re almost done.!Click the link to confirm your email address..</p>'.$html);
        //$this->email->send();

        if($this->email->send()) {
            echo '<script>alert("Email sent successfully")</script>';
            } else {
        $this->email->print_debugger();
        }

        if($ret){
         // $this->db->where('id', $ret);
         // $query=$this->db->get('tbl_user');
        //  $result = $query->result_array();
         // return $result;
          return $this->select1($token);
        }

    }
public function select1($token) {   
    $this->db->select('user_type'); $this->db->from('tbl_user'); 
    $this->db->where('token',$token); 
    return $this->db->get()->row();
}

您可以尝试使用此功能进行插入。请在同一型号中使用此功能。

Answer 2

在控制器中

    public function verify()
    {
        $token= $this->uri->segment(3);
        $email_verification = $this->load->site_model->verifyEmailAddress($token);

        if ($email_verification != 1)
        {
            redirect('site/index');
        }
        else
        {
            $type=$this->load->site_model->select1($token);
            //print_r($type);
            if($type['user_type']=="Employer"){
                redirect('site/emp');
            }
            else{
                redirect('site/seek');
            }

        }
    }

在模型中

    public function verifyEmailAddress($token)
    {
        $data=array('email_verification'=>1);
        $this->db->where('token',$token);
        $this->db->update('tbl_user',$data);
        return 1;

    }
    public function select1($token) 
    {
        $query = $this->db->query("SELECT user_type FROM tbl_user WHERE token = '$token' ");
        $result = $query->result_array();
        return $result;
    }

Answer 3

我知道我可能会迟到1年，但所有答案都只是代码，而stackoverflow答案应该更多地是解决问题的指导。

您收到此错误的原因是因为模型中的 Select1 函数返回了一个对象：$.ajax({ url: "index2.php?id=upload", type: "POST", data: new FormData($("#form")[0]), contentType: false, cache: false, processData:false, dateType:"json", success: function(data) { console.log('names:'+JSON.stringify(data)); $.each(data, function(i,v) { console.log('name:'+v.name); } },

在控制器中，您将返回的值保存为

return $this->db->get()->row();

然后，你有if-else语句，条件是：

$type=$this->load->site_model->select1($token);

您正在尝试使用$type['user_type']=="Employer";的键（索引）获取变量$type，这是一个数组，而您的模型返回该对象。

因此，解决方案是在控制器中从"user_type"更改为$type['user_type']=="Employer"，实际上是指对象的名为“user_type”的属性。

检查if / else中没有工作的codeigniter中的usertype

3 个答案: