PHPでループする方法

PHP どう轮循環
私は想求能力满足量何订单?何么我这样不行??

$order_id = mysql_query("select order_id from myr_order where '".strtotime('-'.$ date.' day')."'
while($t_id = mysql_fetch_array($order_id,MYSQL_ASSOC)) $is_in =1;
$num = 0;
$que = mysql_query("select Goods_sn,order_id from myr_order_goods where order_id = '".$t_id['order_id']."'");
while($gs = mysql_fetch_object($que)){
if(in_array($gs->gt;goods_sn,$sn) && $is_in==1){
$is_in = 1;
}elseif (in_array($gs->goods_sn,$sn) && $is_in==0){
$is_in = 0;
ブレーク;
}
}
if($is_in = = 1){
$num = $num + 1;
}elseif($is_in == 0){
continue;
}
}
echo $num;


------解解案---------
PHP コード

$sql = "select order_id from myr_order where '".strtotime('-'.$date.' day')."' <= add_time および add_time <= '".mktime()."'"; $order_id = mysql_query($sql); $num = 0;// 外面に放り出して一试 while($t_id = mysql_fetch_array($order_id,MYSQL_ASSOC)){ $is_in =1; $que = mysql_query("select Goods_sn,order_id from myr_order_goods where order_id = '".$t_id['order_id']."'"); $gs = mysql_fetch_object($que); if(in_array($gs->goods_sn,$sn) && $is_in){ $is_in = 1; }elseif(in_array($gs->gt;goods_sn,$sn) && !$is_in){ $is_in = 0; 壊す; } if($is_in){ $num = $num + 1; }elseif(!$is_i){ 続く; } } $num をエコーし​​ます。

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