How to make a regular expression for this seemingly simple case?

Question

Is it possible to write a regular expression that works with String replaceAll() to remove zero or more spaces from the end of a line and replace them with a single period (.).

For example, a regex that will make these (non-exhaustive) test cases pass:

String regex = "????";
assertEquals("foo.", "foo".replaceAll(regex, "."));
assertEquals("foo.", "foo ".replaceAll(regex, "."));
assertEquals("foo.", "foo  ".replaceAll(regex, "."));
assertEquals(" foo.", " foo  ".replaceAll(regex, "."));

I know I can do it with replaceFirst(" *$", "."), but I'm wondering if it can be done with replaceAll().

UPDATED TO ADD: The reason for the question is I was writing a data-driven chain of transformations and it only had replace and replaceAll as allowable transformations. I can do it in two replaceAll transformations. I just wondered if it were possible to do it one.

Wiktor Stribiżew · Accepted Answer · 2021-10-20T15:55:42.107

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You can use an alternation based pattern that matches either one or more whitespaces at the end of string or an end of string that has no whitespace right before:

 +$|$(?<! )
\h+$|$(?<!\h)
\s+$|$(?<!\s)

See the regex demo. \h matches horizontal whitespaces, and \s matches any whitespace chars.

Also, $ can be replaced with \z to match the very end of string.

See the Java demo:

String regex = "\\s+\\z|\\z(?<!\\s)";
System.out.println("foo.".equals("foo".replaceAll(regex, ".")));
System.out.println("foo.".equals("foo ".replaceAll(regex, ".")));
System.out.println("foo.".equals("foo  ".replaceAll(regex, ".")));
System.out.println(" foo.".equals(" foo  ".replaceAll(regex, ".")));

All test cases show true as output.

edited Oct 20 '21 at 15:55

answered Oct 20 '21 at 15:49

Wiktor Stribiżew

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Ah, I see, using a negative lookbehind for the second alternative to prevent the double match. – ᴇʟᴇvᴀтᴇ Oct 20 '21 at 15:53
Good point about using `\z` instead of `$`. – ᴇʟᴇvᴀтᴇ Oct 20 '21 at 16:15

sp00m · Answer 2 · 2021-10-20T16:03:42.857

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You could make sure your "list of spaces" is preceded by "not a space":

(?<=\S)\s*$

To cover empty strings too:

(?<=\S|^)\s*$

edited Oct 20 '21 at 16:03

answered Oct 20 '21 at 15:53

sp00m

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That won't work for an empty string. I know that wasn't one of my test cases though! – ᴇʟᴇvᴀтᴇ Oct 20 '21 at 15:57
@ᴇʟᴇvᴀтᴇ Ah yes indeed, I didn't think an empty string should become a single dot, you could still `(?<=\S|^)` if needed, updated. – sp00m Oct 20 '21 at 16:03
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This works and I've upvoted, but the lookbehind approach is slightly shorter. – ᴇʟᴇvᴀтᴇ Oct 20 '21 at 16:13

score 1 · Answer 3 · answered Oct 20 '21 at 19:40

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(?<!\s)\s*$

You could make sure your "list of spaces" is not preceded by "a space".

Instead of: You could make sure your "list of spaces" is preceded by "not a space"

Because (?<=\S) (as character set) doesn't match an empty string, but (?<!\s) does.

Regex101 Demo

answered Oct 20 '21 at 19:40

JGomez

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Is it correct to add the final period when the last _non-space_ character is already a period? – JGomez Oct 20 '21 at 20:02

How to make a regular expression for this seemingly simple case?

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