$seid = isset($seid) ? intval($seid) : 0;
$seid or showmessage('非法参数!', $PHP_SITEURL); $page = isset($page) ? intval($page) : 1;
$pagesize = $HS['pagesize'] ? $HS['pagesize'] : 30;
$offset = ($page-1)*$pagesize;
$id = $seid;$servers = array();
$rs = $db->query("SELECT id,servername FROM ".TABLE_GAMESERVER." WHERE id =$id ");
while($r = $db->fetch_array($rs))
{
$servers[] = $r;
} $sqldb = real_con($id);
$sqldb->select_db(game_fso($id));
$us = $sqldb->get_one("select userid from alluser where loginname = '$_username'");
$r = $sqldb->get_one("SELECT count(*) as number FROM playerbagequipinfo WHERE userid='$us[userid]' AND forgelevel> 3");
$pages = phppages($r['number'], $page, $pagesize);
$exchanges = array();
$result = $sqldb->query("SELECT goodsid,forgelevel,holegem1,holegem2,holegem3 FROM playerbagequipinfo WHERE userid='$us[userid]' AND forgelevel> 3 ORDER BY forgelevel DESC LIMIT $offset,$pagesize");
while($r = $sqldb->fetch_array($result))
{
$r['goodsid'] = getgoodsname($id,$r['goodsid']);
$r['holegem1'] = getgoodsname($id,$r['holegem1']);
$r['holegem2'] = getgoodsname($id,$r['holegem2']);
$r['holegem3'] = getgoodsname($id,$r['holegem3']);
$exchanges[] = $r;
}
function doSub($goodsid,$holegem1,$holegem2,$holegem3)
{
global $db,$_username,$gid,$hid1,$hid2,$hid3;
if($goodsid == '') return "<span class=font06>错误的请求!</span>";
if($holegem1 == '') return "<span class=font06>该装备没有镶嵌宝石!</span>";
if (!check_con($id)) return "<span class=font06>当前分区连接失败,请与管理员联系</span>";
$sqldb = real_con($id);
$sqldb->select_db(game_fso($id));
$us = $sqldb->get_one("select userid from alluser where loginname = '$_username'");
if(!$us) return "<span class=font06>您没有激活该分区</span>";
我想把$servers['id']和$servers['servername']赋给$exchanges['id'],$exchanges['servername'],$exchanges是没有这两个键值的,同时我又不希望破坏$exchanges原有的值。哪位大大能给点建议?或者让$id在dosub里生效也行。谢了
$seid or showmessage('非法参数!', $PHP_SITEURL); $page = isset($page) ? intval($page) : 1;
$pagesize = $HS['pagesize'] ? $HS['pagesize'] : 30;
$offset = ($page-1)*$pagesize;
$id = $seid;$servers = array();
$rs = $db->query("SELECT id,servername FROM ".TABLE_GAMESERVER." WHERE id =$id ");
while($r = $db->fetch_array($rs))
{
$servers[] = $r;
} $sqldb = real_con($id);
$sqldb->select_db(game_fso($id));
$us = $sqldb->get_one("select userid from alluser where loginname = '$_username'");
$r = $sqldb->get_one("SELECT count(*) as number FROM playerbagequipinfo WHERE userid='$us[userid]' AND forgelevel> 3");
$pages = phppages($r['number'], $page, $pagesize);
$exchanges = array();
$result = $sqldb->query("SELECT goodsid,forgelevel,holegem1,holegem2,holegem3 FROM playerbagequipinfo WHERE userid='$us[userid]' AND forgelevel> 3 ORDER BY forgelevel DESC LIMIT $offset,$pagesize");
while($r = $sqldb->fetch_array($result))
{
$r['goodsid'] = getgoodsname($id,$r['goodsid']);
$r['holegem1'] = getgoodsname($id,$r['holegem1']);
$r['holegem2'] = getgoodsname($id,$r['holegem2']);
$r['holegem3'] = getgoodsname($id,$r['holegem3']);
$exchanges[] = $r;
}
function doSub($goodsid,$holegem1,$holegem2,$holegem3)
{
global $db,$_username,$gid,$hid1,$hid2,$hid3;
if($goodsid == '') return "<span class=font06>错误的请求!</span>";
if($holegem1 == '') return "<span class=font06>该装备没有镶嵌宝石!</span>";
if (!check_con($id)) return "<span class=font06>当前分区连接失败,请与管理员联系</span>";
$sqldb = real_con($id);
$sqldb->select_db(game_fso($id));
$us = $sqldb->get_one("select userid from alluser where loginname = '$_username'");
if(!$us) return "<span class=font06>您没有激活该分区</span>";
我想把$servers['id']和$servers['servername']赋给$exchanges['id'],$exchanges['servername'],$exchanges是没有这两个键值的,同时我又不希望破坏$exchanges原有的值。哪位大大能给点建议?或者让$id在dosub里生效也行。谢了
解决方案 »
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$exchanges['servername'] = $servers['servername'];