I want to create a Regular expression to ignore each first and second word of a selected sentence

For example I have this phrase "October 27 New Store Products / October 2022". I want to create a regex that will choose only this part of the phrase ~ "New Store Products / October 2022" and ignore the first date part of the phrase ~ "October 27".

CodePudding user response：

Without knowledge of your true requirements, all we can do is provide best guess, so here is mine;

What you could do, is have something such as the following;

/^\S \s \S \s (.*)$/

What this would do is the following;
From the beginning of the string (^), find one or more non-whitespace chars (\S ), find one or more whitespace chars (\s ) - repeat this again and then use a capture group ((.*)) to get everything else until the end of the string ($).

If you are using JavaScript, you could use this as such;

let sentence = "October 27 New Store Products / October 2022";
let regex = /^\S \s \S \s (.*)$/;
let match = regex.exec(sentence);

if (match) {
  // Ignores the first and second words of the sentence
  console.log(match[1]); // Output: "New Store Products / October 2022" ignoring "October 27"
}

Further explanation of this regex taken from regex101¹ when this is put into the regex bar

/^\S \s \S \s (.*)$/

^ asserts position at start of the string
\S matches any non-whitespace character (equivalent to [^\r\n\t\f\v ])
matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy)
\s matches any whitespace character (equivalent to [\r\n\t\f\v ])
matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy)
\S matches any non-whitespace character (equivalent to [^\r\n\t\f\v ])
matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy)
\s matches any whitespace character (equivalent to [\r\n\t\f\v ])
matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy)
1st Capturing Group (.*)
. matches any character (except for line terminators)
* matches the previous token between zero and unlimited times, as many times as possible, giving back as needed (greedy)
$ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)

CodePudding user response：

You've not provided any information on the context, but does it need to be a regular expression?

String manipulation by searching on spaces might be easier.

For example in PHP:

$string = "October 27 New Store Products / October 2022";
$string_array = explode(' ', $string, 3);
if (array_key_exists(2, $string_array)) echo $string_array[2];

or Excel:

=RIGHT(A1,LEN(A1)-FIND(" ",A1,FIND(" ",A1) 1))