say();?>输出结果:Nciaer">
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关于php承袭的疑惑

时间:2021-07-01 10:21:17 帮助过:22人阅读

关于php继承的疑惑
PHP code
  1. <!--
  2. Code highlighting produced by Actipro CodeHighlighter (freeware)
  3. http://www.CodeHighlighter.com/
  4. -->
  5. <!--?php
  6. class A {
  7. [color=#FF0000]private[/color] $name = "Nciaer";
  8. public function say() {
  9. echo $this ---> name . "<br>";
  10. }
  11. }
  12. class B extends A {
  13. public $name = "Fansa";
  14. }
  15. $b = new B();
  16. $b -> say();
  17. ?>

输出结果:
Nciaer
----------------------------

PHP code
  1. <!--
  2. Code highlighting produced by Actipro CodeHighlighter (freeware)
  3. http://www.CodeHighlighter.com/
  4. -->
  5. <!--?php
  6. class A {
  7. [color=#FF0000]public[/color] $name = "Nciaer";
  8. public function say() {
  9. echo $this ---> name . "<br>";
  10. }
  11. }
  12. class B extends A {
  13. public $name = "Fansa";
  14. }
  15. $b = new B();
  16. $b -> say();
  17. ?>


输出结果:
Fansa


为什么父类变量$name为public的时候,$b调用成员方法say()输出子类的$name;
当父类变量$name为private时,$b调用成员方法say()输出父类的$name;

------解决方案--------------------
private 私有的
怎么能被继承呢?

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