Question

I have this specific line of text and clients submit, and the line automatically breaks at 65 characters and i need to modify my regular expression to look for one more additional line, return carriages or new line characters up until the first punctuation mark such as .,!,?

An example of the line being matched.:

Snippet: It's time again for the Annual State Marching Event, This Saturday at 5pm.

The above line exceeds 65 characters, and sometimes the lines are broken like this

Snippet: It's time again for the Annual State Marching Event, This 
Saturday at 5pm.

Which appears as Event, This\r\n
This is the regexp i use:

preg_match('/Snippet:(\s)?+(.*)\./i', $html_part, $matches) ? $matches[2] : '';

I’ve tried on my won to escape the punctuation marks, but no luck.
Any help would be appreciated. Thanks in advance.

Answer 1

Try the s modifier, like so:

/Snippet:(\s)?+(.*)\./is

If this modifier is set, a dot metacharacter in the pattern matches
all characters, including newlines. Without it, newlines are excluded.
This modifier is equivalent to Perl’s /s modifier. A negative class
such as [^a] always matches a newline character, independent of the
setting of this modifier.

Source

Demo