php - 通过 mysqli_fetch_array 来回浏览

Question

你如何判断案例1是否进入下一个结果案例2是否进入上一个结果非常感谢（我还是有点新，会收到任何帮助）

<?php
$search = mysqli_query($mysqli,"SELECT * FROM `Products` ORDER BY `Products`.`abc_phone` ASC");
while($AnswerP = mysqli_fetch_array($search)){

   /*code...
   ...
   */

   $next = $agi->get_data($FolderToPlay."next_and_oder",7000,1)['result'];
   switch ($next){
    case 1:
      next...;
    break;
    case 2:
        previous...;
    break;
    case 3:
        oder();
    break;
    }
}
?>

标签： phpagi