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我試圖通過特定的「記錄ID」從我的表(tbl_room_image)獲取所有記錄,並在我的視圖中顯示它們與分頁我可以完全限制所有記錄但我的分頁鏈接不工作查看下一條記錄數據。例如,我設置了限制4,所以當頁面加載時,我只能看到前4條記錄不是其他人的幫助。分頁鏈接無法正常工作
這裏是我的控制器:
function addRoomImage($roomid, $limit = '', $offset = 0){
$limit = 4;
$offset = $this->uri->segment(3);
$config['base_url'] = base_url().'config/addRoomImage/'.$roomid;
$config['total_rows'] = $this->config_mdl->count_room_image($roomid);
$config['per_page'] = $limit;
$config['uri_segment'] = 3;
$data['imageData'] = $this->config_mdl->get_roomImage($roomid, $limit, $offset);
$this->pagination->initialize($config);
$data['imageid'] = $this->config_mdl->get_roomById($roomid);
$data['pagination'] = $this->pagination->create_links();
$data['main_content'] = 'config/addRoomImage';
$this->load->view('_base/layout',$data);
}
這裏是我的模型:
function count_room_image($roomid)
{
$this->db->where('room_id', $roomid);
return $this->db->count_all_results('tbl_room_image');
}
function get_roomById($roomid)
{
return $this->db->get_where('tbl_room_info', array('room_id' => $roomid))->result();
}
function get_roomImage($roomid, $limit, $offset)
{
$this->db->select('room_name,room_image');
$this->db->from('tbl_room_image');
$this->db->join('tbl_room_info', 'tbl_room_info.room_id = tbl_room_image.room_id', 'inner');
$this->db->where('tbl_room_image.room_id', $roomid);
$this->db->limit($limit, $offset);
return $this->db->get()->result();
}
首先。 '$ this-> config_mdl-> count_room_image($ roomid);'返回正確的行數。第二。設置'$ config'爲**分頁**,但是這些都不會到'$ this-> config_mdl-> get_roomImage($ roomid,$ limit,$ offset);' 所以我認爲你總是發送默認參數。正確? – 2014-10-20 11:39:57
是的我試圖發送這些參數。 – 2014-10-20 12:28:34
你不需要'$ offset = $ this-> uri-> segment(3);' – seanyt123 2014-10-20 12:30:48