Question

我想创建一个包含搜索表单中的字符串和来自分页

的页面的URL

例如，我的基本网址为example.com/search
显示搜索结果时，它将为example.com/search/?s=keyword，当显示下一页的搜索结果时，第二页的搜索结果为example.com/search/?s=term&p=2，依此类推

如何在CodeIgniter中制作它？

Answer 1

试试这个。这里在网址

中使用了page而不是p

$keyword = trim($this->input->get('s', TRUE));
$this->load->library('pagination');
$config['total_rows'] = $this->db->get('table_name')->num_rows();
$config['per_page'] = 10;
$config['num_links'] = 5;
$config['enable_query_strings'] = TRUE;
$config['use_page_numbers'] = TRUE;
$config['query_string_segment'] = 'page';
$config['page_query_string'] = TRUE;
$config['base_url'] = site_url('search/index/?s=' . $keyword);
$config['full_tag_open'] = '<ul class="pagination">';
$config['full_tag_close'] = '</ul>';
if ($this->input->get('page')) {
    $sgm = (int) trim($this->input->get('page'));
    $segment = $config['per_page'] * ($sgm - 1);
} else {
    $segment = 0;
}

$this->pagination->initialize($config);

// your query
$query = $this->db->select('*')->from('table_name')->limit($config['per_page'], $segment)->get();

现在您的网址与您在问题中提到的网址相同

CodeIgniter搜索输入和分页URL

1 个答案: