我想知道static
关键字在trait中返回什么?它似乎被绑定到trait上,而不是使用它的类上。例如:
trait Example
{
public static $returned;
public static function method()
{
if (!eval(''''.get_called_class().'::$returned;')) {
static::$returned = static::doSomething();
}
return static::$returned;
}
public static function doSomething()
{
return array_merge(static::$rules, ['did', 'something']);
}
}
class Test {}
class Test1 extends Test
{
use Example;
protected static $rules = ['test1', 'rules'];
}
class Test2 extends Test
{
use Example;
protected static $rules = ['test2', 'rules'];
}
// usage
Test1::method();
// returns expected results:
// ['test1', 'rules', 'did', 'something'];
Test2::method();
// returns unexpected results:
// ['test1', 'rules', 'did', 'something'];
// should be:
// ['test2', 'rules', 'did', 'something'];
我可以在method()
方法中使用一些讨厌的eval()
来工作:
public static function method()
{
if (!eval(''''.get_called_class().'::$returned;')) {
static::$returned = static::doSomething();
}
return static::$returned;
}
现在它简单地匹配它对'My'Namespaced'Class::$returned
,但它也很奇怪,因为检查一个静态属性,$returned
,这是定义在trait开始,并正确地绑定到使用它的类。那么为什么static::$returned
不能工作?
那么为什么在这里使用延迟静态绑定呢?
试试这个
trait Example
{
public static $returned;
public static function method()
{
if (!self::$returned) {
self::$returned = self::doSomething();
}
return self::$returned;
}
public static function doSomething()
{
return array_merge(self::$rules, ['did', 'something']);
}
}