用正则表达式组匹配计算？

Question

是否可以使用正则表达式组匹配进行计算？

字符串：

(00) Bananas
...
(02) Apples (red ones)
...
(05) Oranges
...
(11) Some Other Fruit
...

如果每行开头的数字之间的差为 3 或更少，则删除其间的“...”。所以字符串应该像这样返回：

(00) Bananas
(02) Apples (red ones)
(05) Oranges
...
(11) Some Other Fruit

正则表达式：

$match = '/(*ANYCRLF)\((\d+)\) (.+)$
\.{3}
\((\d+)\) (.+)/m';

现在棘手的部分是如何获取匹配项并将一些匹配项添加到类似

的条件中

if($3-$1 >= 3) {
  //replace
}

测试：http://codepad.viper-7.com/f6iI4m

谢谢！

Answer 1

以下是使用 preg_replace_callback() 的方法。

$callback = function ($match) {
    if ($match[3] <= $match[2] + 3) {
        return $match[1];
    } else {
        return $match[0];
    }
};

$newtxt = preg_replace_callback('/(^\((\d+)\).+$)\s+^\.{3}$(?=\s+^\((\d+)\))/m', $callback, $txt);

/(^\((\d+)\).+$)\s+^\.{3}$(?=\s+^\((\d+)\))/m

这是碎片的模式：

(^\((\d+)\).+$)      # subpattern 1, first line; subpattern 2, the number
\s+^\.{3}$           # newline(s) and second line ("...")
(?=\s+^\((\d+)\))    # lookahead that matches another numbered line 
                     # without consuming it; contains subpattern 3, next number

因此，整个模式的匹配是前两行（即编号行和'...'行）。

如果数字差大于3，则替换为$match[0]中的原文（即无变化）。如果差异小于或等于 3，则仅替换为第一行（在 $match1] 中找到）。

Answer 2

您可以使用preg_replace_callback 并使用任何php 代码返回替换字符串，回调接收捕获。但是，对于您的输出，您必须获得重叠匹配以进行替换：

1. 比较(00) Bananas与(02) Apples -> 2-0=2 替换
2. 比较(02) Apples与(05) Oranges -> 5-2=3 替换
3. ...

但由于输入的 (02) Apples 部分已用于上一场比赛，因此您不会第二次拾起它。

编辑：

这是一个基于正则表达式的超前解决方案，归功于 Wiseguy：

$s = "(00) Bananas
...
(02) Apples (red ones)
...
(05) Oranges
...
(11) Some Other Fruit
...";

$match = '/(*ANYCRLF)\((\d+)\) (.+)$
\.{3}
(?=\((\d+)\) (.+))/m';

// php5.3 anonymous function syntax
$s = preg_replace_callback($match, function($m){
    if ($m[3] - $m[1] <= 3) {
        print preg_replace("/[\r\n]+.../", '', $m[0]);
    } else {
        print $m[0];
    }
}, $s);
echo $s;

这是我的第一次拍摄，基于“找到点然后查看上一行/下一行”的逻辑：

$s = "(00) Bananas
...
(02) Apples (red ones)
...
(05) Oranges
...
(11) Some Other Fruit
...
(18) Some Other Fruit
...
(19) Some Other Fruit
...
";

$s = preg_replace("/[\r\n]{2}/", "\n", $s);

$num_pattern = '/^\((?<num>\d+)\)/';
$dots_removed = 0;

preg_match_all('/\.{3}/', $s, $m, PREG_OFFSET_CAPTURE);
foreach ($m[0] as $i => $dots) {
    $offset = $dots[1] - ($dots_removed * 4); // fix offset of changing input

    $prev_line_end = $offset - 2; // -2 since the offset is pointing to the first '.', prev char is "\n"
    $prev_line_start = $prev_line_end; // start the search for the prev line's start from its end
    while ($prev_line_start > 0 && $s[$prev_line_start] != "\n") {
        --$prev_line_start;
    }

    $next_line_start = $offset + strlen($dots[0]) + 1;
    $next_line_end = strpos($s, "\n", $next_line_start);
$next_line_end or $next_line_end = strlen($s);

    $prev_line = trim(substr($s, $prev_line_start, $prev_line_end - $prev_line_start));
    $next_line = trim(substr($s, $next_line_start, $next_line_end - $next_line_start));

    if (!$next_line) {
        break;
    }

    // get the numbers
    preg_match($num_pattern, $prev_line, $prev);
    preg_match($num_pattern, $next_line, $next);

    if (intval($next['num']) - intval($prev['num']) <= 3) {
        // delete the "..." line
        $s = substr_replace($s, '', $offset-1, strlen($dots[0]) + 1);
        ++$dots_removed;
    }
}

print $s;