带有动态属性名称的isset()

Question

isset()当属性名称在变量中时,为什么不起作用？

$Object = new stdClass();
$Object->tst = array('one' => 1, 'two' => 2);

$tst = 'tst'; $one = 'one';
var_dump( $Object, isset( $Object->tst['one'] ), isset( $Object->$tst[ $one ] ) );

输出以下内容:

object(stdClass)#39 (1) {
  ["tst"]=>
  array(2) {
    ["one"]=>
    int(1)
    ["two"]=>
    int(2)
  }
}
bool(true)
bool(false) // was expecting true here..

---

编辑:继续玩弄代码,并发现了

var_dump( $Object->$tst['one'] );

输出通知:

E_NOTICE: Undefined property: stdClass::$t

所以我认为问题在于,在$tst[...]从对象获取属性之前,在"字符串模式"(评估字符串中的第一个字符;在本例中为"t")中评估该部分;

var_dump( $tst, $tst['one'] ); // string(3) "tst" string(1) "t"

---

解决方案:是在变量name($this->{$tst})周围加上大括号,告诉解释器首先检索它的值,然后计算[...]部分:

var_dump( $Object->{$tst}['one'] ); // int(1) yay!

Answer 1

尝试在属性名称周围添加大括号...

isset( $Object->{$tst}[ $one ] );

CodePad.