所以我有一个名为users_online的表,当用户登录我的网站时,该表会自动创建1行,包含date,ip,name,user_id和friend_array(保存所有用户的朋友)

例如,我登录我的网站,并在users_online表中创建行.我想只在网上看到我的朋友,这些朋友都存储在friend_array栏目中(1,5,16,5(那是朋友的身份证号码))..我如何从friends_array列中获取数据并查看这些id中的哪一个被记录目前,这意味着哪些id存在于user_online表中并显示在我的个人资料中？

我希望不要混淆问题……

那就是我的代码..所有商店都在online.PHP文件中：

// Checking wheter the visITor is already marked as being online:
$inDB = MysqL_query("SELECT user_id From users_online WHERE user_id=".$userid);

if(!MysqL_num_rows($inDB))
{
    // Selects some data required to insert into users_online table from users table
    $DB = MysqL_query("SELECT img,fname,friend_array FROM users WHERE user_id=".$userid);
    while($row=MysqL_fetch_assoc($DB))
    {
    $img = $row['img'];
    $fname = $row['fname'];
    $farray = $row['friend_array'];
    }
    MysqL_query("   INSERT INTO users_online (user_id,img,friend_array)
                    VALUES(".$userid.",'".$intIp."','".$img."','".$fname."','".$farray."')");
}
else
{
    // If the visitor is already online,just update the dt value of the row:
    MysqL_query("UPDATE users_online SET dt=Now() WHERE user_id=".$userid);
}


// Counting all the online visitors:
// Thats where i need to work out with friend array..
// I need to display all online friends only
list($totalOnline) = MysqL_fetch_array(MysqL_query("SELECT COUNT(*) FROM users_online"));


// Outputting the number as plain text:
echo $totalOnline;