基于数据库数据筛选具有自动选择的选项的表

Question

我有一个表，我有一个筛选器，目前将按部门筛选（目前只有选项是admin），我希望基于所有部门生成筛选器选项输入数据库，以便在添加新员工并输入新部门时，它将自动显示为过滤器中的一个选项。那可能吗？

<?php 
     if(isset($_GET['aksi']) == 'delete'){ 
      $id = $_GET['id']; 
      $cek = mysqli_query($db, "SELECT * FROM employees WHERE id='$id'"); 
      if(mysqli_num_rows($cek) == 0){ 
       echo '<div class="alert alert-info alert-dismissable"><button type="button" class="close" data-dismiss="alert" aria-hidden="true">&times;</button>Data not found</div>'; 
      }else{ 
       $delete = mysqli_query($db, "DELETE FROM employees WHERE id='$id'"); 
       if($delete){ 
        echo '<div class="alert alert-primary alert-dismissable"><button type="button" class="close" data-dismiss="alert" aria-hidden="true">&times;</button> Employee removed</div>'; 
       }else{ 
        echo '<div class="alert alert-danger alert-dismissable"><button type="button" class="close" data-dismiss="alert" aria-hidden="true">&times;</button> Employee removal failed</div>'; 
       } 
      } 
     } 
     ?> 

     <form class="form-inline" method="get"> 
     <a href="add.php" title="Add" class="btn btn-primary-add btn"><span class="glyphicon glyphicon-plus" aria-hidden="true"></span></a> 
      <div class="form-group"> 
       <select name="filter" class="form-control" onchange="form.submit()"> 
        <option value="0">Filter By Department</option> 
        <?php $filter = (isset($_GET['filter']) ? strtolower($_GET['filter']) : NULL); ?> 
        <option value="Admin" <?php if($filter == 'Admin'){ echo 'selected'; } ?>>Admin</option> 
       </select> 
      </div> 
     </form> 
     <br /> 
     <div class="table-responsive"> 
     <table class="table table-striped table-hover"> 
      <tr> 
       <th>ID</th> 
       <th></th> 
       <th>First Name</th> 
       <th>Last Name</th> 
       <th>D.O.B</th> 
       <th>Telephone</th> 
       <th>E-mail</th> 
       <th>Job Title</th> 
       <th>Site</th> 
       <th>Department</th> 
       <th>Tools</th> 
      </tr> 
      <?php 
      if($filter){ 
       $sql = mysqli_query($db, "SELECT * FROM employees WHERE department='$filter' ORDER BY last_name ASC"); 
      }else{ 
       $sql = mysqli_query($db, "SELECT * FROM employees ORDER BY last_name ASC"); 
      } 
      if(mysqli_num_rows($sql) == 0){ 
       echo '<tr><td colspan="8">Data Tidak Ada.</td></tr>'; 
      }else{ 
       $no = 1; 
       while($row = mysqli_fetch_assoc($sql)){ 
        echo ' 
        <tr> 
         <td>'.$row['id'].'</td> 
         <td><a href="details.php?id='.$row['id'].'"><span class="glyphicon glyphicon-user" aria-hidden="true"></span></td> 
         <td>'.$row['first_name'].'</a></td> 
         <td>'.$row['last_name'].'</td> 
         <td>'.$row['dob'].'</td> 
         <td>'.$row['telephone'].'</td> 
         <td>'.$row['email'].'</td> 
         <td>'.$row['job_title'].'</td> 
         <td>'.$row['site'].'</td> 
         <td>'.$row['department'].'</td> 
         <td> 
          <a href="details.php?id='.$row['id'].'" title="View" class="btn btn-primary-view btn-sm"><span class="glyphicon glyphicon-unchecked" aria-hidden="true"></span></a> 
          <a href="edit.php?id='.$row['id'].'" title="Edit" class="btn btn-primary btn-sm"><span class="glyphicon glyphicon-edit" aria-hidden="true"></span></a> 
          <a href="password.php?id='.$row['id'].'" title="Change Password" data-placement="bottom" data-toggle="tooltip" class="btn btn-warning btn-sm"><span class="glyphicon glyphicon-refresh" aria-hidden="true"></span></a> 
          <a href="index.php?aksi=delete&id='.$row['id'].'" title="Delete" onclick="return confirm(\'Are you sure you want to delete '.$row['first_name'].'?\')" class="btn btn-danger btn-sm"><span class="glyphicon glyphicon-trash" aria-hidden="true"></span></a> 
         </td> 
        </tr> 
        '; 
        $no++; 
       } 
      } 
      ?> 

Answer 1

您应该使用jquery datatable plugin为您的表添加大量的东西。