How to ignore whitespace in a regular expression subject string?

You can stick optional whitespace characters \s* in between every other character in your regex. Although granted, it will get a bit lengthy.

/cats/ -> /c\s*a\s*t\s*s/

While the accepted answer is technically correct, a more practical approach, if possible, is to just strip whitespace out of both the regular expression and the search string.

If you want to search for "my cats", instead of:

myString.match(/m\s*y\s*c\s*a\*st\s*s\s*/g)

Just do:

myString.replace(/\s*/g,"").match(/mycats/g)

Warning: You can't automate this on the regular expression by just replacing all spaces with empty strings because they may occur in a negation or otherwise make your regular expression invalid.

Addressing Steven's comment to Sam Dufel's answer

Thanks, sounds like that's the way to go. But I just realized that I only want the optional whitespace characters if they follow a newline. So for example, "c\n ats" or "ca\n ts" should match. But wouldn't want "c ats" to match if there is no newline. Any ideas on how that might be done?

This should do the trick:

/c(?:\n\s*)?a(?:\n\s*)?t(?:\n\s*)?s/

See this page for all the different variations of 'cats' that this matches.

You can also solve this using conditionals, but they are not supported in the javascript flavor of regex.

You could put \s* inbetween every character in your search string so if you were looking for cat you would use c\s*a\s*t\s*s\s*s

It's long but you could build the string dynamically of course.

You can see it working here: http://www.rubular.com/r/zzWwvppSpE

If you only want to allow spaces, then

\bc *a *t *s\b

should do it. To also allow tabs, use

\bc[ \t]*a[ \t]*t[ \t]*s\b

Remove the \b anchors if you also want to find cats within words like bobcats or catsup.

This approach can be used to automate this (the following exemplary solution is in python, although obviously it can be ported to any language):

you can strip the whitespace beforehand AND save the positions of non-whitespace characters so you can use them later to find out the matched string boundary positions in the original string like the following:

def regex_search_ignore_space(regex, string):
    no_spaces = ''
    char_positions = []

    for pos, char in enumerate(string):
        if re.match(r'\S', char):  # upper \S matches non-whitespace chars
            no_spaces += char
            char_positions.append(pos)

    match = re.search(regex, no_spaces)
    if not match:
        return match

    # match.start() and match.end() are indices of start and end
    # of the found string in the spaceless string
    # (as we have searched in it).
    start = char_positions[match.start()]  # in the original string
    end = char_positions[match.end()]  # in the original string
    matched_string = string[start:end]  # see

    # the match WITH spaces is returned.
    return matched_string

with_spaces = 'a li on and a cat'
print(regex_search_ignore_space('lion', with_spaces))
# prints 'li on'

If you want to go further you can construct the match object and return it instead, so the use of this helper will be more handy.

And the performance of this function can of course also be optimized, this example is just to show the path to a solution.

The accepted answer will not work if and when you are passing a dynamic value (such as "current value" in an array loop) as the regex test value. You would not be able to input the optional white spaces without getting some really ugly regex. Konrad Hoffner's solution is therefore better in such cases as it will strip both the regest and test string of whitespace. The test will be conducted as though both have no whitespace.