Question

在codeigniter中，我使用sqlsrv和分页库来显示数据库中的记录，但是我点击分页链接上的第2页后，记录将显示在当前页面的末尾，而不是显示在新页面中，任何帮助将不胜感激，这是控制器：

public function example1() {
    $config = array();
    $config["base_url"] = base_url() . "index.php/posts/example1";
    $config["total_rows"] = $this->PostModels->record_count();
    $config["per_page"] = 10;
    $config["uri_segment"] = 3;
    $config["num_links"] = 5;

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

    $page = ($this->uri->segment(3)) ? $this->uri->segment(3) : 0;
    $data["results"] = $this->PostModels->
    fetchdata($config["per_page"], $page);
    $data["links"] = $this->pagination->create_links();

    $this->load->view("example1", $data);

以下是模型：

public function record_count() {
return $this->db->count_all("BlawBlaw");
}

public function fetchdata($limit, $start) {
    $this->db->limit($limit, $start);
    $query = $this->db->get("BlawBlaw");

if ($query->num_rows() > 0) {
  foreach ($query->result() as $row) {
    $data[] = $row;
  }
  return $data;
}
return false;
}

＆安培;这是视图：

        <table>
        <tr>
            <th><strong>Post Id</strong></th>
            <th><strong>Post Title</strong></th>
            <th><strong>Post Title</strong></th>
            <th><strong>Post Title</strong></th>
        </tr>
        <?php foreach($results as $data){?>
            <tr>
                <td><?php echo $data->a;?></td>
                <td><?php echo $data->b;?></td>
                <td><?php echo $data->c;?></td>
                <td><?php echo $data->d;?></td>
            </tr>
        <?php }?>
        <p><?php echo $links; ?></p>
    </table>

Answer 1

您可以使用此模型并查看代码中的更改

以下是模型：

function fetch_data($limit, $id) 
   {
       $this->db->select('*')->from('category ')->limit($limit, $id);
        //$this->db->limit($limit);
        //$this->db->where('Cid', $id);
        $query = $this->db->get();
        if ($query->num_rows() > 0) 
        {
        foreach ($query->result() as $row) 
        {
        $data[] = $row;
        }

        return $data;
        }
        return false;


& here is the View:

            <table>
            <tr>
                <th><strong>Post Id</strong></th>
                <th><strong>Post Title</strong></th>
                <th><strong>Post Title</strong></th>
                <th><strong>Post Title</strong></th>
            </tr>
            <?php foreach($results as $data){?>
                <tr>
                    <td><?php echo $data->a;?></td>
                    <td><?php echo $data->b;?></td>
                    <td><?php echo $data->c;?></td>
                    <td><?php echo $data->d;?></td>
                </tr>
            <?php }?>
            <p><?php foreach ($links as $link) {
echo $link; ?></p>
        </table>

CodeIgniter分页：从SQL Server显示记录

1 个答案: