I have a string like s = "abc.def..ghi". I would like to replace the single'.' with two '.'s. However, s.replace(".", "..") yields "abc..def....ghi". How can I get the correct behaviour? The output I'm looking for is s = "abc..def..ghi".
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4 Answers
Replace the dot only when it's surrounded by two characters
String foo = s.replaceAll("(\\w)\\.(\\w)", "$1..$2");
Or as @Thilo commented, only if it's surrounded by two non dots
String foo = s.replaceAll("([^.])\\.([^.])", "$1..$2");
And to replace the single dot with two dots even if the dot is at the beginning/end of the string, use a negative lookahead and lookbehind:
(Example String: .abc.def..ghi. will become ..abc..def..ghi..)
String foo = s.replaceAll("(?<!\\.)\\.(?!\\.)", "..");
2 Comments
Adam
Can you recommend a good "getting started with regex" resource?
baao
I tink regular-expressions.info isn't too bad, though it doesn't look very nice. But you'll find very much there. The lookahead/behind is explained here: regular-expressions.info/lookaround.html @Adam
If you don't know how to do it using regex then use StringTokenizer to split the String and again concatenate with ...
Code:
public static void main(String[] args) {
String s = "abc.def..ghi";
StringTokenizer s2 = new StringTokenizer(s, ".");
StringBuilder sb = new StringBuilder();
while(s2.hasMoreTokens())
{
sb.append(s2.nextToken().toString());
sb.append("..");
}
sb.replace(sb.length()-2, sb.length(), ""); // Just delete an extra .. at the end
System.out.println(sb.toString());
}
I know that the code is big compared to one line regex but I posted it in case if you are having any trouble with regex. If you think it is slower than the accepted answer, I did a start and end time with System.currentTimeMillis() and I got mine faster. I don't know if there are any exceptions to that test.
Anyway, I hope it helps.
7 Comments
Thilo
I don't think this works if you have three consecutive dots (although to be fair, the expected behaviour is underspecified, I assume you'd want to keep them as three dots and not turn them into two dots).
SkrewEverything
@Thilo I tried dots with starting,ending and with multiple dots(>3) in-between. It worked.
SkrewEverything
@Thilo it's
a..bThilo
If we are going with collapsing, the regex solution gets much simpler, too :
replaceAll("\\.+", "..")SkrewEverything
@Thilo Sorry my english understanding is very very bad. I assume you are saying that
a.b is not gonna be converted to a..b? If I'm wrong, please specify a specific string where my code can go wrong( if you don't mind). I think I have checked every possible way to break it and it didn't. |
Use look arounds:
str = str.replaceAll("(?<!\\.)\\.(?!\\.)", "..")(
The regex reads as "a dot, not preceded by a dot, and not followed by a dot". This will handle the edge cases of a dot being at the start/end and/or surrounded by literally any other character.
Because look arounds don't consume input, you don't have to sully yourself with back references in the replacement term.
Comments
s.replace("...","..");
s.replace("....","..");
Depends on input possibilities..
2 Comments
James Real
I mean he needs to use these after the first call to s.replace. In addition, he needs to impose some control on the input stream of such string or else no parser in the world can help him.
Andrew Morton
Please have a look at the answer by baao and consider if you want to delete this answer.