php中的对象问题

Question

我试图习惯于编写面向对象的东西,但我常常陷入困境.我有这个想要用来创建数组的查询.

    $user1 = mysql_fetch_object(mysql_query("SELECT * FROM tbl WHERE username='$oc->user1'"));      
    $user2 = mysql_fetch_object(mysql_query("SELECT * FROM tbl WHERE username='$oc->user2'"));
    $user3 = mysql_fetch_object(mysql_query("SELECT * FROM tbl WHERE username='$oc->user3'"));

我可以通过这样做获取user1的位置:$location = $user1->location,但我想创建一个包含用户所有位置的数组,我试过这个:

        for ($x = 0 ; $x < $users_total; $x++ ){

        $user = "user".($x + 1);            
        $user_location[$x] = $user->location ; //array med stedet alle brukerne befinner seg    

        } 

这应该调用$user1->location,$user2->location,$user3->location到阵列[0] [1]和[3].但相反,我得到了

Notice: Trying to get property of non-object

有什么问题？我不能像这样在字符串中添加数字"user".($x + 1);并调用该位置$user->location.是不是一样的$user1->location？

Answer 1

你想要的$$user->location.查看变量变量.(很多)更好,使用数组.