php preg_replace 如何排除案例

Question

I have the following preg_replace

foreach ($financeTermsArray as $term => $info) {

   $content = preg_replace($term, $info, $content, 1);
}

where the $info variable looks like

'<a class="glossar-word" href="'. get_permalink(). '" data-glossar="'. $excerpt. '">'. get_the_title(). '</a>';

and $term variable like "/\b". get_the_title(). "\b/" "/\b". get_the_title(). "\b/"

I want to exclude from preg_replace if matched word is inside data-glossar html attribute.

How would I do this?

I got it working with the following pattern:

"/data-glossar=\"([^\"]*)(*SKIP)(*F)|\b". get_the_title(). "\b/"

Answer 1

Here is a regex that is constructed with the get_the_title() at runtime.
When the regex is constructed, it will SKIP ALL TAGS in html, including
invisible content, and will only match the '\b'. get_the_title(). '\b' '\b'. get_the_title(). '\b'
as found OUTSIDE tags.

I don't have anyway to test out the resulting regex string, so you should do that
first and see if it looks ok.

Use this on a preg_replace as originally intended.

$term = '~<(?:(?:(?:(script|style|object|embed|applet|noframes|noscript|noembed)(?:\s+(?>"[\S\s]*?"|\'[\S\s]*?\'|(?:(?!/>)[^>])?)+)?\s*>)[\S\s]*?</\1\s*(?=>))|(?:/?[\w:]+\s*/?)|(?:[\w:]+\s+(?:"[\S\s]*?"|\'[\S\s]*?\'|[^>]?)+\s*/?)|\?[\S\s]*?\?|(?:!(?:(?:DOCTYPE[\S\s]*?)|(?:\[CDATA\[[\S\s]*?\]\])|(?:--[\S\s]*?--)|(?:ATTLIST[\S\s]*?)|(?:ENTITY[\S\s]*?)|(?:ELEMENT[\S\s]*?))))>(*SKIP)(*FAIL)|'
        . '\b' . get_the_title() . '\b~';