【发布时间】:2015-08-28 13:09:01
【问题描述】:
我想在 codeigniter 中加入两个表。我想根据登录用户显示数据。我这样做是为了获取用户的会话 ID。它会显示我在标题中提到的错误消息。当我评论“$this->db->where("id",$this->session->userdata['logged_in']['id']);" 这部分它会正常工作。
型号
function get_user(){
$this->db->select("user.firstname,user.lastname,user.address, user.email, user.contact_no, project.location");
$this->db->where("id",$this->session->userdata['logged_in']['id']); //check login user
$this->db->from('user');
$this->db->join('project', 'project.client_id = user.id');
$query = $this->db->get();
return $query->result();
控制器
public function index(){
$data['post'] = $this->profile_model->get_user(); // calling Post model method getPosts()
$this->load->view('user_include/header');
$this->load->view('user_site/profile',$data);
}
查看
<?php if($post) { ?>
<?php foreach($post as $post){?>
<div><?php echo $post->firstname ; echo ' '; echo $post->lastname;?></div> <br>
<div><?php echo $post->address ; echo ' '?></div><br>
<div><?php echo $post->contact_no ; echo ' '?></div> <br>
<div><?php echo $post->email ; echo ' '?></div> <br>
<div><?php echo $post->location ; echo ' '?></div> <br>
<?php }
} else {
?>
</div>
<div clospan="4" align="center">No records found to display</div>
【问题讨论】:
-
试试
$this->db->where("user.id",$this->session->userdata['logged_in']['id']); -
也许别名 where ID...
where("user.id",$this->se看来项目和用户可能都有表列ID -
感谢萨蒂。它工作正常
标签: mysql codeigniter