controller:
public function view()
{
$seid=$this->session->userdata('user_id');
$data['result']=$this->search_model->getall($seid);
$this->load->view('pagination_view',$data);
}
model:
public function getall($seid)
{
$this->db->select('uid','name','details','img');
$this->db->where("email",$seid);
$q=$this->db->get('tbl_reg');
return $q->result();
}
view:
<?php foreach($result as $row){?>
<?php <?php echo "username:".$row['name'];die();?>
<br/>
<?php echo "email:" .$row['email'];?>
<br/>
<?php echo "user id:" .$row['uid'];?>
<?php echo "Details:".$row['details'];?>
<img src="<?php echo $row['img'] ;}?>" style="width:104px;height:142px"; />
when i logged in to the home page , did not display the current logged in user details. what is the problem with this code .Please provide solution for this problem?
If you want to get single row from database please use $q->row_object()
or $q->row_array()
. Then you don't need to use foreach on view. Please see below example:-
Your Modal
public function getall($seid)
{
$this->db->select('uid','name','details','img');
$this->db->where("email",$seid);
$q=$this->db->get('tbl_reg');
return $q->row_array();
}
On your View
<?php echo "username:".$row['name']; ?>
<br/>
<?php echo "email:" .$row['email'];?>
<br/>
<?php echo "user id:" .$row['uid'];?>
<?php echo "Details:".$row['details'];?>
<img src="<?php echo $row['img'] ;?>" style="width:104px;height:142px"; />
modify your model to use the following:
public function getall($seid)
{
$this->db->select('uid','name','details','img');
$this->db->where("email",$seid);
$q=$this->db->get('tbl_reg');
return $q->result_array();
}
In controller, add the user details in the following manner,
public function view()
{
$seid=$this->session->userdata('user_id');
$data['result']=$this->search_model->getall($seid);
$newdata = array(
'username' => <USERNAME>,
'email' => <MAILID>,
...........
'logged_in' => TRUE
);
$this->session->set_userdata($newdata);
$this->load->view('pagination_view',$data);
}
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