有人可以帮我在提交表单后如何从输入 name="$photo_id" 中获取值。在下一页上应该是 $photo_id = $_GET['photo_id'] ......
$picture = mysql_query( "SELECT * FROM gallery_photos where photo_category = ".$cid." ");
while($row2 = mysql_fetch_array($picture)){
$photo_id = $row2["photo_id"];
$photo_filename = $row2["photo_filename"];
$photo_caption = $row2["photo_caption"];
$photo_category = $row2["photo_category"];
echo "<ul style='float:left; list-style:none; '>";
echo "<li><img src='".$images_dir."/tb_".$photo_filename."' border='0' alt='".$photo_caption."' /><br />";
echo "<span><input name='$photo_id' type='text' value='$photo_caption' /></li></span>";
echo "</ul>";
}
谢谢 :)