有人可以帮助我如何才能从输入名称=“$ photo_id”作为下面的状态提交表单后的值...应该是$ photo_id = $ _GET ['photo_id “]下一页...如何从输入html格式使用php获得价值
$picture = mysql_query("SELECT * FROM gallery_photos where photo_category = ".$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>";
}
THX :)
这是什么形式? – Juice