Php Regex：如何匹配重复的模式

Question

给出以下文字

bond0:       0       0    0    0    0     0          0         0        0       0    0    0    0     0       0          0
eth0: 11329920252 12554462    0    0    0     0          0      3561 13072970332 12899522    0    0    0     0       0          0

我需要捕获列值。我想到了这些行：

Regex: `(\w+):(?:\s+(\d+))+`
Php: `preg_match_all('/(\w+):(?:\s+(\d+))+/sim', $data, $regs)

但不幸的是，它只捕获第一列。

Array
(
    [0] => Array
        (
            [0] => dummy0:       0       0    0    0    0     0          0         0        0       0    0    0    0     0       0          0
            [1] => bond0:       0       0    0    0    0     0          0         0        0       0    0    0    0     0       0          0
            [2] => eth0: 11329920252 12554462    0    0    0     0          0      3561 13072970332 12899522    0    0    0     0       0          0
            [3] => ip6tnl0:       0       0    0    0    0     0          0         0        0       0    0    0    0     0       0          0
            [4] => lo: 51675995  100695    0    0    0     0          0         0 51675995  100695    0    0    0     0       0          0
            [5] => sit0:       0       0    0    0    0     0          0         0        0       0    0    0    0     0       0          0
            [6] => tunl0:       0       0    0    0    0     0          0         0        0       0    0    0    0     0       0          0
        )

    [1] => Array
        (
            [0] => 0
            [1] => 0
            [2] => 0
            [3] => 0
            [4] => 0
            [5] => 0
            [6] => 0
        )

)

有什么建议吗？谢谢 `

====编辑==== 明确一点：我知道我可以 preg_match 搜索 \d+ 值或将整个字符串分成几行并在每一行上运行explode，但我对正则表达式解决方案感兴趣，我将第一列作为第一个成员结果数组（实际上忘记在问题的初稿中放置捕获括号），以及带有数据的后续列，每一行都放在它的专用数组中......

Answer 1

为什么要使用preg_match 或preg_match_all？

$results = array();
foreach (preg_split("/\r\n|\r|\n/", $data) as $line)
{
    list($key, $values) = explode(":", $line);
    $results[$key] = preg_split("/\s/", trim($values));
}

只要每行不超过一个:，这应该可以工作。在我看来，这也是写这个的最短和最快的方法。

Answer 2

给你：

$data = explode("\n", $data);
$out = array();
foreach ($data as $d) {
    preg_match_all('/\s(\d+)/', $d, $matches);

使$matches[0] 等于匹配数组。然后你想将它添加到行数组中：

    $out[] = $matches[0];
}

您现在有一个锯齿状的行和列数组。所以，要参考第二行第四列，你可以去$out[1][3]。

Answer 3

我知道您正在寻找preg_match 解决方案，但这是以防您找不到任何有用的答案

<?php
$val = "bond0:       0       0    0    0    0     0          0         0        0       0    0    0    0     0       0          0
eth0: 11329920252 12554462    0    0    0     0          0      3561 13072970332 12899522    0    0    0     0       0          0";
$arr1 = explode("\n",$val);
foreach ($arr1 as $value) {
    $exp = explode(":",$value);
    $ex = preg_replace('/\s+/', ' ',trim($exp[1]));
    $arr[$exp[0]] = explode(" ",$ex);
}
var_dump($arr);
?>

结果：

array (size=2)
  'bond0' => 
    array (size=17)
      0 => string '' (length=0)
      1 => string '0' (length=1)
      2 => string '0' (length=1)
      3 => string '0' (length=1)
      4 => string '0' (length=1)
      5 => string '0' (length=1)
      6 => string '0' (length=1)
      7 => string '0' (length=1)
      8 => string '0' (length=1)
      9 => string '0' (length=1)
      10 => string '0' (length=1)
      11 => string '0' (length=1)
      12 => string '0' (length=1)
      13 => string '0' (length=1)
      14 => string '0' (length=1)
      15 => string '0' (length=1)
      16 => string '0' (length=1)
  'eth0' => 
    array (size=17)
      0 => string '' (length=0)
      1 => string '11329920252' (length=11)
      2 => string '12554462' (length=8)
      3 => string '0' (length=1)
      4 => string '0' (length=1)
      5 => string '0' (length=1)
      6 => string '0' (length=1)
      7 => string '0' (length=1)
      8 => string '3561' (length=4)
      9 => string '13072970332' (length=11)
      10 => string '12899522' (length=8)
      11 => string '0' (length=1)
      12 => string '0' (length=1)
      13 => string '0' (length=1)
      14 => string '0' (length=1)
      15 => string '0' (length=1)
      16 => string '0' (length=1)

Answer 4

你可以这样做：

$subject = <<<LOD
dummy0: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
bond0:  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
eth0: 11329920252 12554462 0 0 0 0 0 3561 13072970332 12899522 0 0 0 0 0 0
ip6tnl0: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
lo: 51675995 100695 0 0 0 0 0 0 51675995 100695 0 0 0 0 0 0
sit0: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
tunl0: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
LOD;

$pattern = '~^(?<item>\w+):|\G\h+(?<value>\d+)~m';
preg_match_all($pattern, $subject, $matches, PREG_SET_ORDER);
$i=-1;
foreach($matches as $match) {
    if (isset($match['value']))
        $result[$i]['values'][] = $match['value'];
    else {
        $i++;
        $result[$i]['item'] = $match['item']; }
}
print_r($result);

您将获得您在 EDIT 中描述的格式。

图案细节：

~               # pattern delimiter
^               # anchor for the line start (in multiline mode)
(?<item>\w+)    # named capture "item"
:       
|               # OR
\G              # force the match to be contigous from precedent
\h+             #
(?<value>\d+)   # named capture "value"
~m              # pattern delimiter, m modifier for multiline mode