如何在同一页面中保留GET后的值？

Question

我的页面名称是sym.php，我的问题是当我点击下面的时

<a href=sym.php?poscon=Face>Face</a>
   <?php
     echo '<br><td><input type="checkbox" '.$strIsChecked.' title ="'.$otherspec.'" name="sym[]"  onclick="javascript: submit()" value ="'.$spec.'"></td>';
     echo '<td align="">'.$spec.'</td>';
     if (!empty($_POST['sym'])) { 
         foreach($_POST['sym'] as $condition) {
             $condition = mysqli_real_escape_string($link,$condition);
             $p = mysqli_query($link,"select *   from  $tablename where PossibleCondition = '". $condition ."'");  
             echo "<b> ".$condition." </b> <br>" ;
             while ($r = mysqli_fetch_array($p)) {
                 $spec = $r["Specialization"];
                 $spec1 = $r["Specialization1"];   
             }    
   ?>   

点击顶部的href后该值应该返回我该怎么办？我会做什么功能。提前谢谢。

<a href="hgapp.php?doc=<?php echo $spec ?>"> <u><?php echo $spec ?></u></a>
<a href="hgapp.php?doc=<?php echo $spec1 ?>"><u><?php echo $spec1 ?></u></a>
<a href="hgapp.php?doc=<?php echo $spec2 ?>"><u><?php echo $spec2 ?></a></u>
<a href="hgapp.php?doc=<?php echo $spec3 ?>"> <u><?php echo $spec3 ?></a></u>
<?php } } ?>