关于PHP上传文件名的获取

上传文件的代码：
<?php
                  $sql_image="select * from `image` where achrive='1' and productId='".$pId."'";
                  $query_image=mysql_query($sql_image);
                  $count_image=mysql_num_rows($query_image);
                  $i=1;
                  while($result_image=mysql_fetch_array($query_image)){
                  if($count_image>0){
                  ?><?php echo $result_image['path'];?>
   <input type="hidden" value="<?php echo $result_image['imageId']; ?>" name="image_id_<?php echo $i; ?>" />
  <input type="hidden" value="<?php echo $result_image['path']; ?>" name="h_image_<?php echo $i;?>" />
   <img src="../img/110504170234lvotpvxkmhym.jpg" style="width:60px;height:40px;" name="e_image_<?php echo $i;?>" />
     <input style="text-align:left;vertical-align:top;" type="file" name="t_image_<?php echo $i;?>" />//主要看这行       
<input style="text-align:left;vertical-align:top;" type="reset" value="Cancel" /> <input style="text-align:left;vertical-align:top;" type="reset" value="Delete" onClick="javascript:window.location.href='del_image.php?imageId=<?php echo $result_image['imageId'];?>'" />
                                    <br />
                                    <?php
                                        $i++;
                                    }else{                                    }
                                }
                         ?>
<input style="text-align:left;vertical-align:top;" type="file" name="t_image_<?php echo $i;?>" />//主要看这行如何获取文件名，因为name="t_image_<?php echo $i;?> name是动态的，我不会做了，求助各位大侠们。