Question

我遇到了这个问题，我有一个名为person的表，其中包含字段id，name和person_room。我也有这个名为room的表，其中包括roomid，name和capacity字段。我想从桌面的现场容量中调用数据。但它显示资源ID＃7

这是我的代码：

$check_room_capacity = "SELECT capacity from room WHERE roomid = '".$oroom."'";
$check_user_capacity = mysql_query($check_room_capacity);
$rows = mysql_fetch_array($check_user_capacity);

$check_room = "SELECT id from person WHERE person_room = '".$oroom."'";
$check_user_room = mysql_query($check_room);

    if(mysql_num_rows($check_user_room) => $check_user_capacity){
        echo "room is loaded please try another room";
    }

//then code here to insert it if the number of person 
//is less than the room capacity

我试图回复这个东西$ check_user_capacity来知道它的值，但它显示了资源ID＃7。你们有什么想法吗？

另一件事是，为什么我的PHP不接受这个操作＆＃39; =＆gt;＆＃39;和＆＃39;＆lt; =＆＃39;它总是用红色加下划线？

Answer 1

mysql_query（）语句返回指向结果集的资源指针，而不是数据本身。您需要使用mysql_fetch_array（）来检索表中的实际数据。

$old_ip = mysql_query("SELECT current_ip FROM members WHERE id='$id'"); 
$row = mysql_fetch_array($old_ip);
$result_old_ip = $row['current_ip'];

Answer 2

试试这个：

$check_room_capacity = "SELECT capacity from room WHERE roomid = '".$oroom."'";
    $check_user_capacity = mysql_query($check_room_capacity);
    $rows = mysql_fetch_array($check_user_capacity);

    $check_room = "SELECT id from person WHERE person_room = '".$oroom."'";
    $check_user_room = mysql_query($check_room);

        if(mysql_num_rows($check_user_room) >= $rows['capacity']){
            echo "room is loaded please try another room";
        }

如何从数据库中获取数字值？

2 个答案: