获取实例的静态属性
如果我在PHP中有一个实例，那么获取该实例的静态属性（'类变量'）的最简单方法是什么？
这个
$classvars=get_class_vars(get_class($thing));
$property=$classvars['property'];
$thing::property
$thing->property
没有找到相关结果
已邀请:
8 个回复
岭取
$ property当然必须定义为和。蔬谢窗港
class Person { public static $name = 'Joe'; public function iam() { echo 'My name is ' . self::$name; } } $me = new Person(); $me->iam(); // displays "My name is Joe"芦歉竭皑
我想就是这样。信藉乒
标记）访问它们class X { public static $var = 13; } echo X::$var;体悉
博沮富教全
<?php namespace Foo; class Bar { public static $baz=1; //... public function __toString() { return self::class; } } echo Bar::$baz; // returns 1 $bar = new Bar(); echo $bar::$baz; // returns 1<?php namespace Foo; class Bar { public static $baz=1; public $class=self::class; //... } $bar = new Bar(); echo $bar->class::$baz; // returns 1陈獭
因为你发现它的两条线太长，你还有其他选择： 1.写一个吸气剂<?php class C { static $staticvar = "STATIC"; function getTheStaticVar() { return self::$staticvar; } } $instance = new C(); echo $instance->getTheStaticVar();<?php class C { static $staticvar = "STATIC"; function getStatic($staticname) { return self::$$staticname; } } $instance = new C(); echo $instance->getStatic('staticvar');class C { static $staticvar = "STATIC"; function __get($staticname) { return self::$$staticname; } } $instance = new C(); echo $instance->staticvar;掀辟髓观粟
class testClass { public static $property = "property value"; public static $property2 = "property value 2"; } echo testClass::$property; echo testClass::property2;