Php preg_match_所有项目的数组，然后针对每个匹配

我正在编辑一些Interspire电子邮件代码。目前，该程序在发送电子邮件之前会浏览电子邮件的HTML，并查找“a href”代码以替换链接。我希望它也通过并获得formaction=并替换它们中当前没有的URL。我想我可以使用下面这篇文章中的正则表达式：

但是我在思考如何处理阵列时遇到了一些困难。仅执行“a href=”的当前代码如下：

    preg_match_all('%<a.+(href\s*=\s*(["\']?[^>"\']+?))\s*.+>%isU', $this->body['h'], $matches);
    $links_to_replace = $matches[2];
    $link_locations = $matches[1];

    arsort($link_locations);
    reset($links_to_replace);
    reset($link_locations);

    foreach ($link_locations as $tlinkid => $url) {
        // so we know whether we need to put quotes around the replaced url or not.
        $singles = false;
        $doubles = false;

        // make sure the quotes are matched up.
        // ie there is either 2 singles or 2 doubles.
        $quote_check = substr_count($url, "'");
        if (($quote_check % 2) != 0) {
        ...

现在还不成立，是吗？有可能做到我想的吗？或者我是否需要编写另一个函数来处理与“a href”分开的“forms action=”

a建议：

$pattern = <<<'LOD'
~
(?|            # branch reset feature: allows to have the same named
               # capturing group in an alternation. ("type" here)
    <a\s           # the link case
    (?>  # atomic group: possible content before the "href" attribute
        [^h>]++        # all that is not a "h" or the end of the tag ">"
      |
        \Bh++          # all "h" not preceded by a word boundary
      |
        h(?!ref\s*+=)  # all "h" not followed by "ref=" or "ref    ="
    )*+  # repeat the atomic group zero or more times.
    (?<type> href )
  | #### OR ####
    <form\s        # the form case
    (?> # possible content before the "action" attribute. (same principle)
        [^a>]++
      |
        \Ba++
      |
        a(?!ction\s*+=)
    )*+
    (?<type> action )
)
\s*+ = \s*+     # optional spaces before and after the "=" sign
\K              # resets all on the left from match result
(?<quote> ["']?+ )
(?<url> [^\s"'>]*+ )
\g{quote}       # backreference to the "quote" named capture (", ', empty)
~xi
LOD;

---

看起来您试图在preg_match_all中传递多个参数。为什么不使用preg_replace来替换action=URL或DOM格式？如果我假设正确，您希望替换数组元素中的URL，最好使用preg_replace_callback。如果您希望执行替换，为什么不使用preg_replace或preg_replace_callback？我对PHP不是很精通，所以不知道最好的方法或唯一的方法。所以，如果我想添加一个函数，通常我只需要离开已经存在的东西，并尝试添加到它。你是说preg_match_一切都不管用？我假设他们使用这个函数是有原因的，这取决于preg_replace，就像我说的，如果你想找到所有form action=URL，然后从数组元素中替换这些URL，那么最好使用preg_replace_回调函数来执行。如果不是这样，那么请更清楚地说明你到底想做什么=

$links_to_replace = $matches[2];
$link_locations = $matches[1];

$pattern = <<<'LOD'
~
(?|            # branch reset feature: allows to have the same named
               # capturing group in an alternation. ("type" here)
    <a\s           # the link case
    (?>  # atomic group: possible content before the "href" attribute
        [^h>]++        # all that is not a "h" or the end of the tag ">"
      |
        \Bh++          # all "h" not preceded by a word boundary
      |
        h(?!ref\s*+=)  # all "h" not followed by "ref=" or "ref    ="
    )*+  # repeat the atomic group zero or more times.
    (?<type> href )
  | #### OR ####
    <form\s        # the form case
    (?> # possible content before the "action" attribute. (same principle)
        [^a>]++
      |
        \Ba++
      |
        a(?!ction\s*+=)
    )*+
    (?<type> action )
)
\s*+ = \s*+     # optional spaces before and after the "=" sign
\K              # resets all on the left from match result
(?<quote> ["']?+ )
(?<url> [^\s"'>]*+ )
\g{quote}       # backreference to the "quote" named capture (", ', empty)
~xi
LOD;

$html = preg_replace_callback($pattern,
    function ($m) {
        $url = $m['url'];
        $type = lowercase($m['type']);
        $quote = $m['quote'];
        // make what you want with the url, type and quotes
        return $quote . $url . $quote;        
    }, $html);