Php 父类中可获取的laravel
我使用的是Laravel 5,我有模型课Php 父类中可获取的laravel,php,laravel,oop,laravel-5,Php,Laravel,Oop,Laravel 5,我使用的是Laravel 5,我有模型课 class Apartment extends Property { protected $table = 'apartments'; } 和父类 abstract class Property extends Model { protected $table = ''; public function doSomthing () { echo $table = $this->getTab
class Apartment extends Property
{
protected $table = 'apartments';
}
和父类
abstract class Property extends Model
{
protected $table = '';
public function doSomthing ()
{
echo $table = $this->getTable(); //$this->table
}
}
如何在父类中读取$table公寓或获取表名
abstract class Property extends Model
{
protected $table = '';
public function doSomthing ()
{
echo $table = $this->getTable(); //$this->table
}
}
$this->getTable;和$this->table and self::$table不起作用除了这不是一个模型之外,这个变量是什么。。这表明您将代码放在了错误的部分。您可能应该从理解开始,.Non-static method App\Property::doSomthing不应该被静态调用,在我将静态添加到公共静态函数doSomething之后,我得到了1。doSomthing可以像平常一样调用$this->doSomthing;2.如果doSomthing必须是一个静态函数,请尝试将$this->table作为参数传递给doSomthing$table{…},您不能在静态函数中使用$this是的,我已经制作了这个doSomthing$table,但我要求提供更好的解决方案,非常感谢u
class Apartment extends Property
{
protected $table = 'apartments';
public function __construct()
{
parent::__construct($this->table);
}
}
abstract class Property extends Model
{
protected $table = '';
function __construct($var)
{
$this->table = $var; //$this->table should be 'apartments'
}
}