【发布时间】:2016-02-19 23:26:19
【问题描述】:
这些是我编写的一些示例代码
<?php
/**************** code block 1 *****************/
class Database1
{
public $rows;
public function __construct()
{
}
public function assingrow()
{
$this->rows=5;
}
}
$database1 = new Database1();
$database1->assingrow();
echo $database1->rows;
//no error
/**************** code block 2 *****************/
class Database2
{
protected $rows;//
public function __construct()
{
}
public function assingrow()
{
$this->rows=5;
}
}
$database2 = new Database2();
$database2->assingrow();
echo $database2->rows;
//error message
//Fatal error: Cannot access protected property Database2::$rows in E:\xampp\htdocs\pdo\4.php on line 46
/**************** code block 3 *****************/
class Database3
{
public $rows;
public function __construct()
{
}
protected function assingrow()////
{
$this->rows=5;
}
}
$database3 = new Database3();
$database3->assingrow();
echo $database3->rows;
//error message
//Fatal error: Call to protected method Database3::assingrow() from context '' in E:\xampp\htdocs\pdo\4.php on line 68
/**************** code block 4 *****************/
class Database4
{
public $rows;
protected function __construct()
{
}
public function assingrow()
{
$this->rows=5;
}
}
$database4 = new Database4();
$database4->assingrow();
echo $database4->rows;
//error message
//Fatal error: Call to protected Database4::__construct() from invalid context in E:\xampp\htdocs\pdo\4.php on line 91
谁能解释一下为什么会这样
- 为什么不能在代码块 2 中为受保护的属性赋值
- 为什么不能在代码块 3 中使用受保护的方法为公共属性赋值
- 为什么不能在代码块 4 中保护构造
【问题讨论】:
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请阅读documentation,如果之后有问题,请写下