Convert String like *italic* to italic but do not convert if line breaks are with-CodePudding

I have the following Regex in my PHP code:

// markers for italic set *Text*
if (substr_count($line, '*')>=2)
{
    $line = preg_replace('#\*{1}(.*?)\*{1}#', '<i>$1</i>', $line);
}

which works great.

However, when a $line holds a  , e.g.

*This is my text<br>* Some other text

Then the regex still considers the text and transforms it to:

<i>This is my text<br></i> Some other text

The goal is to not translate the text if a   is encountered. How to do that with a Regex - using a so called "negative lookahead" or how can the existing Regex be changed?

Note: Strings like *This is my text* Some other text And again *italic* END should still be considered and transformed.

Idea: Or should I explode the $line and then iterate over the results with the regex?!

CodePudding user response：

Using match-what-you-don't-want and discard technique, you may use this regex in PHP (PCRE):

\*[^*]*<br>\*(*SKIP)(*F)|\*([^*]*)\*

and replace with $1

RegEx Demo

PHP code:

$r = preg_replace('/\*[^*]*<br>\*(*SKIP)(*F)|\*([^*]*)\*/'), 
        "<i>$1</i>", $input);

Explanation:

\*: Match a *
[^*]*: Match 0 or more non-* characters
 : Match  
\*: Match closing *
(*SKIP)(*F): PCRE verbs to discard and skip this match
|: OR
\*([^*]*)\*: Match string enclosed by *s

CodePudding user response：

You can replace matches of the regular expression

\*(?:(?!<br>)[^*]) \*

with

'<i>$0</i>'

where $0 holds the matched string.

Demo

The regular expression can be broken down as follows.

\*         # match '*'
(?:        # begin a non-capture group
  (?!<br>) # negative lookahead asserts that next four chars are not '<br>'
  [^*]     # match any char other than '*'
)          # end non-capture group and execute one or more times
\*         # match '*'