PHP中文网2017-04-17 18:00:14
This should be a syntax issue, and it has nothing to do with generics
can only be written as class name.class, which is the syntax agreed by java
PHP中文网2017-04-17 18:00:14
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获取当前对象的三种方式:
1.类名.calss
2.Class.forName("类的完整路径");
3.对象.getClass()
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