分頁codeigniter始終在第2頁開始，並且活動鏈接始終在第2頁

Question

我是codeigniter的新手，我通過從類別ID獲取查詢結果使用codeigniter分頁，但頁面總是從第2頁開始。 如何修復分頁以設置頁碼1的開始頁面？

public function kategori($kat=''){ 

$config['base_url'] = base_url().'index.php/lowongan/kategori/'.$kat; 
$data['db']=$this->query->katPostingan($kat);  
$config['total_rows'] = $data['db']->num_rows(); 
$config['per_page'] = 1;  
$config['uri_segment'] = 4; 
$this->pagination->initialize($config); 
$data2['paging'] = $this->pagination->create_links(); 

$page = ($this->uri->segment($config['uri_segment'])) ? $this->uri->segment($config['uri_segment']) : 0; 
$data2['records'] = $this->query->katPos($kat, $config['per_page'], $page); 

$this->load->view('user/hal_kategori',$data2); 
    } 

和我的模型

public function katPos($kat='',$limit,$start){ 
$this->db->where("id_kat", $kat); 
$this->db->limit($limit,$start); 
    $query = $this->db->get("showKat"); 
    return $query; 
} 

    public function katPostingan($kat=''){ 
      $this->db->where("id_kat", $kat); 
      $query = $this->db->get("showKat"); 
     return $query; 
    } 

Answer 1

嘗試的解決方案不同的方式，

控制器（指數（））

 $this->load->library('pagination'); 
     $config['base_url'] = site_url().'classname/index'; 
     $config['per_page'] = $this->config->item('pagination_number'); 
     $config['uri_segment'] = 4; 
     $data['count']=$this->example_model->count(); 
     $config['total_rows'] =$data['count']; 
     $data['classname']=$this->example_model->getall($this->config->item('pagination_number'),$this->uri->segment(4)); 
     $this->pagination->initialize($config); 
     $data['page_link']=$this->pagination->create_links(); 


     $this->load->view('users/classname.php',$data); 

模型

function getall($num,$offset) 
{ 
    $this->db->from('table_name'); 
    $this->db->limit($num,$offset); 
    $query = $this->db->get(); 
    return $query->result_array(); 
} 
function count() 
{ 
    $this->db->from('table_name'); 
    $query = $this->db->get(); 
    $rowcount = $query->num_rows(); 
    return($rowcount); 
} 

查看

<div class="pagination"> <?php echo $page_link;?> </div>