Question

我有以下php代码：

with open(path_in + data_file) as csvfile:
    read_it = list(csv.reader(csvfile, delimiter=','))

结果是

<?php

class ParentClass {

    public $table_name;

    function __construct() {
        $this->table_name = strtolower(__CLASS__);
    }

}

class ChildClass extends ParentClass {

    function __construct() {
        parent::__construct();
        //And I also want to put here other codes
    }

}

$parent = new ParentClass();
$child = new ChildClass();

echo $parent->table_name . "<br />" . $child->table_name;

?>

然而我希望它是

parentclass
parentclass

我如何实现它？后期静态绑定可以解决我的问题，但当然不能静态调用。 http://php.net/manual/en/language.oop5.late-static-bindings.php

Answer 1

这适用于PHP5.5及更高版本：

class ParentClass 
{
    public $table_name;

    public function __construct() 
    {
        $this->table_name = strtolower(static::class);
    }
}

class ChildClass extends ParentClass 
{
    public function __construct() 
    {
        parent::__construct();     
        // ...
    }
}

$parent = new ParentClass();
$child = new ChildClass();

echo $parent->table_name . "<br />" . $child->table_name;

供参考，请参阅http://php.net/manual/en/language.oop5.basic.php#language.oop5.basic.class.class。

Answer 2

使用static::class使用后期静态绑定获取类的名称：

class ParentClass {

    public $table_name;

    function __construct() {
        $this->table_name = strtolower(static::class);
    }

}

这将产生所需的

parentclass
childclass

Answer 3

<?php

class ParentClass {

    public $table_name;

function __construct() {
    $this->table_name = strtolower(__CLASS__);
}

}

class ChildClass extends ParentClass {

function __construct() {

    parent::__construct();
    $this->table_name = strtolower(__CLASS__);
    //And I also want to put here other codes
}

}

$ parent = new ParentClass（）; $ child = new ChildClass（）;

echo $ parent-＆gt; table_name。＆＃34;
＆＃34; 。 $儿童安全＆GT;表名;

＆GT;

静态地从子类调用__construct（？）

3 个答案: