PHP 怎麼輪循

來源:互聯網
上載者:User
PHP 如何輪循
我是想求能滿足多少個訂單?為什麼我這樣不行??

$order_id = mysql_query("select order_id from myr_order where '".strtotime('-'.$date.' day')."' <= add_time and add_time <= '".mktime()."'");
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($gs->goods_sn,$sn) && $is_in==1){
$is_in = 1;
}elseif(in_array($gs->goods_sn,$sn) && $is_in==0){
$is_in = 0;
break;
}
}
if($is_in == 1){
$num = $num + 1;
}elseif($is_in == 0){
continue;
}
}
echo $num;

------解決方案--------------------
PHP code
$sql = "select order_id from myr_order where '".strtotime('-'.$date.' day')."' <= add_time and add_time <= '".mktime()."'";$order_id = mysql_query($sql);$num = 0;// 放到外面來試一試while($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->goods_sn,$sn) && !$is_in){        $is_in = 0;        break;    }    if($is_in){        $num = $num + 1;    }elseif(!$is_i){        continue;    }}  echo $num;
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