Question

我正在编写一个类来清理通过ajax调用传递给PHP的字符串，当我将一个字符串传递给这个类时，它工作正常，但是将数组作为引用传递，它将无法工作。

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
)

我缺少什么，或者是否无法在PHP中嵌套引用传递？

Answer 1

您的代码不会修改$array，而是会修改$value。

有几种解决方法，一种是foreach ($array as &$value)，另一种是在循环内修改$array[$key]。

Answer 2

你丢失了foreach中的引用。将其改为此，它将起作用：

foreach( $array as $key => &$value ) {

PHP传递数组作为参考

2 个答案: