如果未使用PHP找到带有默认值的RegEx分组

时间:2019-04-15 20:01:34

标签: php regex

这是一个示例字符串:

$text = 'foo (20/50) bar () baz (11/30)';

我需要的输出是这样:

$items = array(
   array(
      "title" => "foo",
      "number" => 20
   ),
   array(
      "title" => "bar",
      "number" => 0
   )
   array(
      "title" => "baz",
      "number" => 11
   )
);

尝试

到目前为止,我一直在使用

$matches_title = array();
$matches_number = array();
preg_match_all('!([^\s]+)!',$text,$matches_title);
preg_match_all('!(?<=\()(\d+)(?=\/)!',$text,$matches_number);

然后遍历匹配项以捕获值。显然,当括号之一为空时,这将不起作用,因为两个数组的长度都不同。

我该如何解决这个问题?

2 个答案:

答案 0 :(得分:3)

您可以使用以下php代码:

$text = 'foo (20/50) bar () baz (11/30)';
preg_match_all('~([\w-]+)\h*\((\d*)(?=[/)])~', $text, $m);

$items = array();
foreach($m[1] as $i => $v) {
   $n = $m[2][$i];
   $items[] = array( "title" => $v, "number" => (empty($n) ? 0: $n) );
}
print_r($items);

输出:

Array
(
    [0] => Array
        (
            [title] => foo
            [number] => 20
        )

    [1] => Array
        (
            [title] => bar
            [number] => 0
        )

    [2] => Array
        (
            [title] => baz
            [number] => 11
        )

)

答案 1 :(得分:1)

另一种选择是使用function getHubspotData(url) { console.log("URL: " + url); return fetch(url) .then((resp) => resp.json()) // Transform the data into json .then(function(data) { console.log(data); return 2; // set resolved value of promise }).then(function (x) { console.log(x); // outputs 2 }); } 来匹配字符串的格式,以在上一个匹配项的末尾声明位置并使用命名的捕获组:

\G

Regex demo | Php demo

例如:

\G(?<title>\S+)\h+\((?:(?<number>\d+)/\d+)?\)(?:\s|$)

结果:

$str = "foo (20/50) bar () baz (11/30)";
$pattern = '~\G(?<title>\S+)\h+\((?:(?<number>\d+)/\d+)?\)(?:\s|$)~';
preg_match_all($pattern, $str, $matches, PREG_SET_ORDER, 0);
$matches = array_reduce($matches, function($carry, $item) {
    $carry[] = [
        "title" => $item["title"],
        "number" => array_key_exists("number", $item) ? $item["number"] : 0

    ];
    return $carry;
});


print_r($matches);