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异为3或更小，则删除其中的“...”。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抓住比赛并添加一些像

这样的条件

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

测试：http://codepad.viper-7.com/f6iI4m

谢谢！

Answer 1

您可以使用preg_replace_callback()来完成此操作。

$callback = function ($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）。

如果数字差异大于3，请替换为$match[0]中的原始文本（即无更改）。如果差异小于或等于3，则仅替换为第一行（在$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：

比较(00) Bananas与(02) Apples - ＆gt; 2-0=2 替换
比较(02) Apples与(05) Oranges - ＆gt; 5-2=3 替换
...

但由于输入的(02) Apples部分已用于上一场比赛，因此第二次不会被选中。

编辑：

这是一个基于正则表达式的解决方案，具有前瞻性，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;

这是我的第一次采取，基于逻辑“找到点，然后看到上一行/下一行”：

$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;

用正则表达式组匹配计算？

2 个答案:

编辑：