Php 如何从对象'；使用变量的字段？

F.e。 我有object$游戏，有3个截图 $game->screen1，$game->screen2，$game->screen3

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我需要像$game['screen'.$num]这样获取数据，但在对象中。

您可以使用数组进行此操作：

$game->screen = array($screen1, $screen2, $screen3);

$game->screen[$num];

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您可以这样做：

$game->screen2 = 'this is screen2';
$num = 2;
echo $game->{'screen' . $num}; // "this is screen2"

$game->screen3 = 'this is screen3';
$num = 3;
echo $game->{'screen' . $num}; // "this is screen3"