wrap words in string with regex

Question

This is the string

(code)

Pivot: 96.75<br />Our preference: Long positions above 96.75 with targets @ 97.8 &amp; 98.25 in extension.<br />Alternative scenario: Below 96.75 look for further downside with 96.35 &amp; 95.9 as targets.<br />Comment the pair has broken above its resistance and should post further advance.<br />

(text)

"Pivot: 96.75
Our preference: Long positions above 96.75 with targets @ 97.8 & 98.25 in extension.
Alternative scenario: Below 96.75 look for further downside with 96.35 & 95.9 as targets.
Comment the pair has broken above its resistance and should post further advance.
"

the result should be

(code)

<b>Pivot</b>: 96.75<br /><b>Our preference</b>: Long positions above 96.75 with targets @ 97.8 &amp; 98.25 in extension.<br /><b>Alternative scenario</b>: Below 96.75 look for further downside with 96.35 &amp; 95.9 as targets.<br />Comment the pair has broken above its resistance and should post further advance.<br />

(text)
Pivot: 96.75
Our preference: Long positions above 96.75 with targets @ 97.8 & 98.25 in extension.
Alternative scenario: Below 96.75 look for further downside with 96.35 & 95.9 as targets.
Comment the pair has broken above its resistance and should post further advance.

The porpuse:
Wrap all the words before : sign.

I've tried this regex: ((\A )|(<br />))(?P<G>[^:]*):, but its working only on python environment. I need this in PHP:

$pattern = '/((\A)|(<br\s\/>))(?P<G>[^:]*):/';
$description = preg_replace($pattern, '<b>$1</b>', $description);

Thanks.

Answer 1

I should start by saying that HTML operations are better done with a proper parser such as DOMDocument. This particular problem is straightforward, so regular expressions may work without too much hocus pocus, but be warned :)

You can use look-around assertions; this frees you from having to restore the neighbouring strings during the replacement:

echo preg_replace('/(?<=^|<br \/>)[^:]+(?=:)/m', '<b>$0</b>', $str);

Demo

First, the look-behind assertion matches either the start of each line or a preceding <br />. Then, any characters except the colon are matched; the look-ahead assertion makes sure it's followed by a colon.

The /m modifier is used to make ^ match the start of each line as opposed to \A which always matches the start of the subject string.

Answer 2

This preg_replace should do the trick:

preg_replace('#(^|<br ?/>)([^:]+):#m','$1<b>$2</b>:',$input)

PHP Fiddle - Run (F9)

Answer 3

The most "general" and least regex-expensive way to do this that I could come up with was this:

$parts = explode('<br', $str);//don't include space and `/`, as tags may vary
$formatted = '';
foreach($parts as $part)
{
    $formatted .= preg_replace('/^\s*[\/>]{0,2}\s*([^:]+:)/', '<b>$1</b>',$part).'<br/>';
}
echo $formatted;

Or:

$formatted = array();
foreach($parts as $part)
{
    $formatted[] = preg_replace('/^\s*[\/>]{0,2}\s*([^:]+:)/', '<b>$1</b>',$part);
}
echo implode('<br/>', $formatted);

Tested with, and gotten this as output

Pivot: 96.75
Our preference: Long positions above 96.75 with targets @ 97.8 & 98.25 in extension.
Alternative scenario: Below 96.75 look for further downside with 96.35 & 95.9 as targets.
Comment the pair has broken above its resistance and should post further advance.

That being said, I do find this bit of data weird, and, if I were you, I'd consider str_replace or preg_replace-ing all breaks with PHP_EOL:

$str = preg_replace('/\<\s*br\s*\/?\s*\>/i', PHP_EOL, $str);//allow for any form of break tag

And then, your string looks exactly like the data I had to parse, and got the regex for that here:

$str = preg_replace(...);
$formatted = preg_replace('/^([^:\n\\]++)\s{0,}:((\n(?![^\n:\\]++\s{0,}:)|.)*+)/','<b>$1:</b>$2<br/>', $str);