第一次使用php和代码点火器,我想显示的数据不会显示在文本字段中,我是否遗漏了代码?
查看
Username: <input type="text" class="form-control" id="username" name="username" placeholder="" disabled="disable" value="<?php echo $result['username']; ?>" > <br>
Password: <input type="text" class="form-control" id="password" name="password" placeholder=""disabled="disable" value="<?php echo $result['password']; ?>"><br>
Rank : <input type="text" class="form-control" id="rank" name="rank" placeholder=""disabled="disable" value="<?php echo $result['title']; ?>"><br>
模型
Class profile_model extends CI_Model {
public function my_data()
{
$username = $this->session->userdata('$username');
$data = array();
$this->db->select('*');
$this->db->from('user');
$this->db->where('username', $username);
$query = $this->db->get();
//this will return multiple rows or object of arrays
//return $query->result();
// you need to send only single row
return $query->row();
}
}
控制器
public function profile() {
// in data array key name should be same which you will pass to view
$this->load->model('profile_model');
$data['result'] = $this->profile_model->my_data();
$this->load->view('profile_view', $data);
}
}
答案 0 :(得分:1)
不需要在模型函数中使用$data=array()
,因为$query->row();
返回对象并使用以下函数而不是您在模型中编写的函数:
public function my_data()
{
$username = $this->session->userdata('$username');
$this->db->select('*');
$this->db->from('user');
$this->db->where('username', $username);
$query = $this->db->get();
return $query->row();
}
并在视图中编写以下代码:
value="<?php echo $result->username; ?>"
答案 1 :(得分:0)
$query->row()
返回一个对象,而不是一个数组。请在您的视图中尝试此操作:
value="<?php echo $result->username; ?>"
答案 2 :(得分:0)
尝试使用row_array()
public function my_data()
{
$username = $this->session->userdata('$username');
$data = array();
$this->db->select('*');
$this->db->from('user');
$this->db->where('username', $username);
$query = $this->db->get();
return $query->row_array();
}