Question

我有一些使用PHP中的USE函数为php 5.3.0编写的代码

有人可以帮我改变这个为5.2.9工作吗？

$available  = array_filter($objects, function ($object) use ($week) { 
    return !in_array($object, $week);
});

感谢您的帮助

Answer 1

不太好，但这将是一个等效的实现。

class MyWeekFilter {
    protected $_week;

    public function __construct($week) {
        $this->_week = $week;
    }

    public function filter($object) {
        return !in_array($object, $this->_week);
    }
}

$filter    = new MyWeekFilter($week);
$available = array_filter($objects, array($filter, 'filter'));

Answer 2

作者代码之间是否有任何区别

$available = array_filter($objects, function ($object) use ($week) { 
    return !in_array($object, $week);
});

和

$available = array_diff($objects, $week);

Answer 3

$available  = array_filter($objects, create_function('$object', '
    $week = '.var_export($week,true).';
    return !in_array($object, $week);
'));

Answer 4

试试这个：

$week = array(...); // defined and instantiated before...

function callback($object) {
    return !in_array($object, $week);
}
$available  = array_filter($objects, "callback");

PHP 5.3.0 USE关键字 - 如何在5.2中反向移植？

4 个答案: