Question

假设我从某些源代码获得此输出，但我无法访问原始PHP创建的数组：

Array
(
    [products] => Array
        (
            [name] => Arduino Nano Version 3.0 mit ATMEGA328P
            [id] => 10005
        )

    [listings] => Array
        (
            [category] => 
            [title] => This is the first line
This is the second line
            [subtitle] => This is the first subtitle
This is the second subtitle
            [price] => 24.95
            [quantity] => 
            [stock] => 
            [shipping_method] => Slow and cheap
            [condition] => New
            [defects] => 
        )

    [table_count] => 2
    [tables] => Array
        (
            [0] => products
            [1] => listings
        )

)

现在我想输入这些数据并让算法重新创建它正在打印的原始数组，这样我就可以将它用于我自己的应用程序。

目前我正在考虑一个sub_str()和正则表达式语句，它们可以提取数据并进行适当的放置。在我进一步讨论之前，是否有一种更简单的方法，通过已经编写的代码或php插件，我已经在那里为我做了这个？

Answer 1

function print_r_reverse($in) {
    $lines = explode("\n", trim($in));
    if (trim($lines[0]) != 'Array') {
        // bottomed out to something that isn't an array
        return $in;
    } else {
        // this is an array, lets parse it
        if (preg_match("/(\s{5,})\(/", $lines[1], $match)) {
            // this is a tested array/recursive call to this function
            // take a set of spaces off the beginning
            $spaces = $match[1];
            $spaces_length = strlen($spaces);
            $lines_total = count($lines);
            for ($i = 0; $i < $lines_total; $i++) {
                if (substr($lines[$i], 0, $spaces_length) == $spaces) {
                    $lines[$i] = substr($lines[$i], $spaces_length);
                }
            }
        }
        array_shift($lines); // Array
        array_shift($lines); // (
        array_pop($lines); // )
        $in = implode("\n", $lines);
        // make sure we only match stuff with 4 preceding spaces (stuff for this array and not a nested one)
        preg_match_all("/^\s{4}\[(.+?)\] \=\> /m", $in, $matches, PREG_OFFSET_CAPTURE | PREG_SET_ORDER);
        $pos = array();
        $previous_key = '';
        $in_length = strlen($in);
        // store the following in $pos:
        // array with key = key of the parsed array's item
        // value = array(start position in $in, $end position in $in)
        foreach ($matches as $match) {
            $key = $match[1][0];
            $start = $match[0][1] + strlen($match[0][0]);
            $pos[$key] = array($start, $in_length);
            if ($previous_key != '') $pos[$previous_key][1] = $match[0][1] - 1;
            $previous_key = $key;
        }
        $ret = array();
        foreach ($pos as $key => $where) {
            // recursively see if the parsed out value is an array too
            $ret[$key] = print_r_reverse(substr($in, $where[0], $where[1] - $where[0]));
        }
        return $ret;
    }
}

不是我的代码，请在评论中找到：print_r '马特'是老板

Answer 2

我还为您的问题制定了解决方案，因为它是一个非常有用的工具，可以在Stack Overflow上重新创建问题。我的来源位于：https://github.com/etalon/aprp

我做的有点不同：在你的字符串中你不需要换行符，所以我走过了字符。一个不利条件：如果数组值中有括号或括号，则无效。

从print_r输出重新创建原始PHP数组

2 个答案: