Regex that fail when there excessive whitespaces between words

Question

I'm trying to make a regex in Java that does not match excessive whitespaces between words, and whitespaces at the beginning and end.

I have done this regex so far

^[\\S].*[\\S]$

That fail when there are whitespaces at the beginning and end of a line

But how about excessive whitespaces between words?

I want this line to fail:

"test    test"

But not this:

"test test"

I tried this:

^[\\S].*(?![\\s]{2,}).*[\\S]$

But it didn't work.

Why don't you just trim the string before sending it to your method? — Luiggi Mendoza, Jun 09 '15 at 16:37
Do you want excessive whitepsace strings to fail, or do you want to trim the whitespace within lines so that it does not fail? — woemler, Jun 09 '15 at 17:41
@willOEM The regex should fail when the String starts with whitespace/ends with whitespace/has 2 or more whitespaces between words. — Jonas P., Jun 09 '15 at 18:01

score 0 · Answer 1 · answered Jun 09 '15 at 16:40

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For failure case just check for:

\s{2,}

i.e. a whitespace 2 or more times. If there is a match then fail the verification.

answered Jun 09 '15 at 16:40

anubhava

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score 0 · Answer 2 · answered Jun 09 '15 at 16:42

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This should check before and after and more than two spaces

[^\s]([ ]{2,})[^\s]

answered Jun 09 '15 at 16:42

guy_sensei

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score 0 · Accepted Answer · edited May 23 '17 at 10:26

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After reading this answer https://stackoverflow.com/a/1240365/2136936

I came up with the following regex, which is exactly what I wanted:

^[\\S](?!.*\\s{2,}).*[\\S]$

One thing I don't understand though, is why it doesn't work this way: ^[\\S].*(?!\\s{2,}).*[\\S]$

edited May 23 '17 at 10:26

Community

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answered Jun 11 '15 at 16:47

Jonas P.

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Your second alternative does not work because it requires that at *at least one position* in the string there are not two whitespaces (`.*` outside of the negation), while the first one requires that nowhere are two whitespaces. – mihi Jun 11 '15 at 17:28

score 0 · Answer 4 · answered Jun 11 '15 at 17:25

You can use the following regex for it (assuming you do not want to match the empty string):

"^\\S++(\\s\\S++)*+$"

First a nonzero amount of non-whitespaces, and then multiple (possible zero) repetitions of a single whitespace followed by multiple (non-zero) non-whitespaces.

Instead of the non-backtracking ++ and *+ operators you can also use (with same result) the normal + and * operators; note that the performance of the non-backtracking operators will be a lot better in case the string is long (several kilobytes).

Regex that fail when there excessive whitespaces between words

4 Answers4