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我到处搜索和搜索,但我无法找到明确的答案在这个问题上。如何实施与Kohana 2.3.x分页搜索
我想做一个网页搜索和分页结果(并按标题排序)。
请参阅原型http://i55.tinypic.com/2dlrqbs.png
我想,如果一个用户指定为搜索条件“A”,包含“A”显示所有名称。我的问题是我怎样才能把导航链接的字符串:?name = a。
如果我不发回搜索条件,单击下一页将显示所有记录。
我看了很多帖子关于这个问题,我还不知道怎么办呢
控制器代码(草案)
function listall()
{
$limit = 2 ;
$orderby = 'u.id';
$direction = 'asc';
$name = '';
if ($_POST)
{
$name = $this->input->post('name');
}
if ($_GET)
{
$name = $this->input->post('name');
if ($this->input->get('orderby'))
list($orderby, $direction) = explode(':', $this->input->get('orderby'));
}
$view = new view('character/listall');
$db = Database::instance();
$sql = "select c.id id, c.lastactiontime, c.name character_name, k.name kingdom_name, k.image kingdom_image, from_unixtime(u.last_login, '%d-%m-%y') last_login, u.nationality
from characters c, kingdoms k, users u
where 1=1 and
c.kingdom_id = k.id and
c.user_id = u.id
" ;
if ($name != '')
$sql .= "and c.name like '%" . $name . "%'" ;
$characters = $db->query($sql);
//$characters = ORM::factory("character")->orderby($orderby, $direction)->find_all();
$this->pagination = new Pagination(array(
'base_url'=>'character/listall',
'uri_segment'=>'listall',
'style'=>'digg',
'query_string' => 'page',
'total_items'=>$characters->count(),
'items_per_page'=>$limit));
//$characters = ORM::factory("character")->orderby($orderby, $direction)->find_all($limit, $this->pagination->sql_offset);
$sql .= " order by $orderby $direction ";
$sql .= " limit $limit offset " . $this->pagination->sql_offset ;
kohana::log('debug', $sql);
$characters = $db->query($sql);
$playersinfo = Character_Model::getplayersinfo();
$view->playersinfo = $playersinfo;
$view->pagination = $this->pagination;
$view->characters = $characters;
$this->template->content = $view;
}
感谢
你的代码应工作? AFAIR,分页会自动添加当前查询字符串。 – biakaveron 2010-12-16 13:41:19