0

可能重复:
通过字符串获取 PHP 类属性

有没有更清洁的方法来做到这一点?它不编译

$object1->name = 'my_name';
$object1->address = 'address';
$object2->somefield->($object1->name) = $object1;

只有当我分配给另一个变量 $object1->name 时它才有效

$object1->name = 'my_name';
$object1->address = 'address';
$temp = $object1->name;
$object2->somefield->$temp = $object1;
4

1 回答 1

9

试试这个:

$object2->somefield->{$object1->name} = $object1;
于 2012-12-11T15:41:39.850 回答