Question

我从数据库和数据中成功选择回来。我只是不确定如何访问数组的元素，如用户名或名称等。这是输出：

数组（[0] =＆gt; stdClass对象（[id] =＆gt; 9 [fullname] =＆gt;Ø®Ø§Ù“Ø¯ Ø§Ù“ØºØ§Ù...Ø¯ÙŠ[membership] =＆gt;免费[用户名] =＆gt; KKKKK [密码] =＆gt; 16d7a4fca7442dda3ad93c9a726597e4 [email] =＆gt; someone@gmail.com [about] =＆gt; [city] =＆gt; riyadh [profilepic] =＆gt; /home2/sdds/public_html/uploads/Red_velvet_cupcakes_with_roses.jpgRed_velvet_cupcakes_with_roses.jpg [mobile] =＆gt; [电话] =＆gt; [地址] =＆gt; [gallery_link] =＆gt; ））

这是返回上述内容的模型中的代码：

public function login()
{
    $this->db->select('*');
    $this->db->from('members');
    $this->db->where('username', $this->input->post('username'));
    $this->db->where('password', md5($this->input->post('password')));
    $query = $this->db->get();
    return $query->result();
}

那么该怎么办？提前谢谢

Answer 1

请检查一下：

//model function in your model    
public function login()
    {
        $this->db->select('*');
        $this->db->from('members');
        $this->db->where('username', $this->input->post('username'));
        $this->db->where('password', md5($this->input->post('password')));
        $query = $this->db->get();
        return $query->row();
    }
   // in controller  
    $logindata = $this->YOUR_MODEL_NAME->login();
// access like this

    echo $logindata->fullname;

参考：https://www.codeigniter.com/user_guide/database/results.html

Answer 2

打桩 - 也要确定并确认你的模型方法中是否有任何东西

 // check how many records returned with num_rows()
 if ($query->num_rows() == 1) {

       // since there is just one record, use row()
       // when 2 or more records, use result() 
       return $query->row();
    }
    else
    { return false; }

在你的控制器中你会立即知道是否有任何回复并且您可以显示适当的视图。

  if(! $logindata = $this->login_model->login() ) { 
   // NO login data came back 
   //  show appropriate view

  else {
 // pass to data if you need it in the view 
  $data['logindata'] = $logindata ;
   // call and show the view  

 }

codeigniter返回对象

2 个答案: