PHP传递数组作为参考

Jonathan Jones 发表于 Dev

乔纳森·琼斯

我正在编写一个类来清理通过ajax调用传递给PHP的字符串，当我将字符串传递给此类时，它可以正常工作，但将数组作为引用传递，将无法工作。

class Sanitize {

    public static function clean (&$str) {
        self::start($str);
    }

    public static function cleanArray (&$array) {
        if (self::arrayCheck($array)) {
            foreach ($array as $key => $value) {
                if (self::arrayCheck($value)) {
                    self::cleanArray($value);
                } else {
                    self::clean($value);
                }
            }
        } else {
            throw new Exception ('An array was not provided. Please try using clean() instead of cleanArray()');
        }
    }

    private static function start (&$str) {
        $str .= '_cleaned';
    }

    private static function arrayCheck ($array) {
        return (is_array($array) && !empty($array));
    }
}

测试代码：

$array = array(
    'one' => 'one',
    'two' => 'two',
    'three' => 'three',
    'four' => 'four'
);
echo print_r($array, true) . PHP_EOL;
Sanitize::cleanArray($array);
echo print_r($array, true) . PHP_EOL;

输出：

Array
(
    [one] => one
    [two] => two
    [three] => three
    [four] => four
)

Array
(
    [one] => one
    [two] => two
    [three] => three
    [four] => four
)

是否有我缺少的东西，还是无法在PHP中嵌套引用传递？

埃里克

您会在foreach中丢失参考。将其更改为此，它将起作用：