我有分页和codeigniter的问题。我有一个来自女巫的quick_searh视图我将信息提交给索引控制器函数并设置分页并调用quick_search方法来获取我想要的数据。它只是不起作用。我花了5个多小时重写这些方法,甚至从quick_search开始,然后转到索引函数,但没有任何效果,请帮忙。
public function index(){
// search parameters config
$lawyer_name = $this->input->post('lawyer_name');
$kanzlei = $this->input->post('kanzlei');
$area_of_expertise = $this->input->post('area_of_expertise');
$post_code = $this->input->post('post_code');
$city = $this->input->post('city');
$result = $this->quick_search(
$this->uri->segment(3),
$lawyer_name,
$kanzlei,
$area_of_expertise,
$post_code,
$city);
if(isset($result)){
// pagination config
$this->load->library('pagination');
$this->load->library('table');
$config['total_rows'] = count($result);
$config['base_url'] = 'http://localhost/anwalt/index.php/search/index';
$config['per_page'] = 5;
$config['num_links'] = 5;
$this->pagination->initialize($config);
$data['search_result_array'] = $result;
$data['main_content'] = 'pages/quick_search_results';
$this->load->view('templates/home_body_content', $data);
}
}
quick_search功能:
public function quick_search($offset, $lawyer_name, $kanzlei, $area_of_expertise, $post_code, $city){
// no input in the quick search
if( empty($lawyer_name) && empty($kanzlei) && empty($area_of_expertise)
&& empty($post_code) && empty($city))
{
$result = 'nothing';
} else {
$this->load->model('quick_search_model');
$result = $this->quick_search_model->get_search_results(
$offset,
$lawyer_name,
$kanzlei,
$area_of_expertise,
$post_code,
$city
);
}
return $result;
}
sql是这样的:
$sql = "SELECT users.user_id, users.canonical_name, first_name, last_name, city, phone_number, kanzlei
from users
inner join user_normal_aos
on users.user_id = user_normal_aos.user_id
inner join normal_areas_of_expertise
on user_normal_aos.normal_areas_of_expertise_id = normal_areas_of_expertise.normal_areas_of_expertise_id
where ".implode(" AND ", $where);
if(empty($offset)){
$offset = 0;
}
$sql = $sql." LIMIT ".$offset.", 4";
显示数据,但我没有看到那里的分页..甚至当我想更改分段的URL时,它说我没有任何数据。
视图如下:
<h1>Quick search results</h1>
<?php
if($search_result_array == "nothing"){
echo "<h3>You havent inputed anything</h3>";
} else {
echo $this->table->generate($search_result_array);
}
echo $this->pagination->create_links();
答案 0 :(得分:1)
根据您的搜索变量,您可以使用:
$lawyer_name = $this->input->post('lawyer_name');
$kanzlei = $this->input->post('kanzlei');
$area_of_expertise = $this->input->post('area_of_expertise');
$post_code = $this->input->post('post_code');
$city = $this->input->post('city');
/*pagination start*/
$this->load->library('pagination');
$config['base_url'] = base_url().'index.php/index/lawyer/'.$lawyer_name.'/kanzlei/'.$kanzlei.'/area_of_expertise/'.$area_of_expertise.'/post_code/'.$city.'/page/';
$config['total_rows'] = $this->model->count_all_results(); ###implement this function to count all the vodeos as per the search variables, just use the same function as "quick_search" but without the limit clause
$config['per_page'] = count($result);;
$config['uri_segment'] = 10;
$config['next_link'] = 'Next';
$config['prev_link'] = 'Prev';
$config['cur_tag_open'] = '<span class="active_page">';
$config['cur_tag_close'] = '</span>';
$this->pagination->initialize($config);
/*pagination end*/
答案 1 :(得分:0)
你不能使用$ this-&gt; pagination-&gt; create_links();视野中的方法。
使用$ data ['pagination'] = $ this-&gt; pagination-&gt; create_links();在加载视图之前的控制器中 并在视图中回显$ pagination
希望这会对你有所帮助。