我如何分頁這些數據？

Question

你會如何分頁？ 數據由下面的「getInstance」查詢調用，然後在動態表中列出一些名爲「pages」whis的記錄。

我想能夠顯示每個查詢只10「頁面」和分頁他們向前

 <table width="100%" border="0" cellpadding="0" cellspacing="0" class="std"> 

     <?php 


     $counter = 1; 
     $userID = PageDB::getInstance()->get_user_id_by_name($_SESSION['user']); 
     $result = PageDB::getInstance()->get_pages_by_campaign_id($campaignID); 
     $i=0; 
      while ($row = mysqli_fetch_array($result)): 
      $style = ""; 
       if($i%2==0) 
{ 
    $style = ' style="background-color: #EFEFEF"'; 
} 
echo "<tr".$style.">"; 
      echo "<td>&nbsp;</td>"; 
      echo "<td></td>"; 


      //The loop is left open 
      ?> 
      <td> 
      <form style="display:none;"></form> 
      <div id="Pagination" class="pagination"> </div> 
    <div id="Searchresult"> </div> 
       <form name="editPage" action="editPage.php" method="GET"> 
        <input type="hidden" name="pageID" value="<?php echo $pageID = $row['pid']; ?>" /> 
        <input type="submit" name="editPage" value="<?php echo "Page " . $counter; ?>" style="background:none;border:none; cursor:pointer"/> 
       </form> 

      </td> 
      <td> 
       <form name="deletePage" action="deletePage.php" method="POST"> 
        <input type="hidden" name="pageID" value="<?php echo $pageID = $row['pid']?>"/> 
        <input type="submit" name="deletePage" value="Delete" onclick = "javascript: return confirm('Delete Page <?php echo $counter ?> ?');"/> 
       </form> 
      </td>    
      <?php 
      $pageID = $row['pid']; 
      $counter++; 
      $i++; 

      echo "</tr>\n"; 
     endwhile; 
     mysqli_free_result($result); 
     ?> 
    </table> 

的查詢REFFERENCE該呼叫中的用戶ID，以調用從corellating頁面上，這2個公共職能DB：

public function get_user_id_by_name($name) { 
    $name = $this->real_escape_string($name); 
    $user = $this->query("SELECT id FROM users WHERE name = '" . $name . "'"); 

    if ($user->num_rows > 0){ 
     $row = $user->fetch_row(); 
     return $row[0]; 
    } else 
     return null; 
} 

public function get_pages_by_campaign_id($campaignID) { 
    $campaignID = $this->real_escape_string($campaignID); 
    return $this->query("SELECT pid , campaignid FROM pages , campaigns WHERE campaignid = campaigns.id"); 
} 

任何人都可以看到解決方案嗎？謝謝！

Answer 1

您可以在MySQL中使用LIMIT查詢。

SELECT pid , campaignid 
FROM pages , campaigns 
WHERE campaignid = campaigns.id 
LIMIT first_record_number, how_many_records 

PEAR Pager可以幫助你獲得當前頁碼等

$pager_options = array(
    'mode'  => 'Sliding', 
    'perPage' => 10, 
    'delta'  => 2, 
    'totalItems' => $total_count, 
); 

$pager = Pager::factory($pager_options); 

list($from, $to) = $pager->getOffsetByPageId(); 
//TODO: get and display data from database 
//first_record_number = $from-1 
//how_many_records = 10 

//show the links 
echo $pager->links; 

我的建議是使用virualization庫數據庫，例如通信。 AdoDB。有方法：

$rs = $conn->SelectLimit("SELECT pid , campaignid 
FROM pages , campaigns 
WHERE campaignid = campaigns.id", how_many_records,first_record_number); 
$arr1 = $rs->GetArray(); 

它也可以做輸入數據的過濾，歌廳所有結果作爲陣列和許多其他有用的東西。