String.split on dot doesn't work as expected

Question

I have a string = 12.05.2014

And I am trying to use split by "." but it is returning empty array for some reason.

System.out.println(newDate);
System.out.println(newDate.length());
String [] dateCells = newDate.split(".");

StringBuilder sBuilder = new StringBuilder();
sBuilder.append(dateCells[2]).append(dateCells[0]).append(dateCells[1]);
System.out.println(sBuilder.toString());

The output is:

12.05.2014
10
            //empty line

The split method takes a _regular expression_ so you need to escape the dot i.e. `split("\\.")`. — Ian Roberts, May 14 '14 at 23:21

Jeroen Vannevel · Answer 1 · 2014-05-14T23:23:04.803

4

Your condition is

for(int i = dateCells.length; i < 0; i--){

It starts at 3 and runs while your counter is less than 0. Which it never is.

Obvious solution: either use i > 0 or change it up to loop from 0 to 3.

Additionally: don't forget to escape the dot which is a special regex character. Escaping is done through a backslash (\) but since that backslash is also a special character, you have to escape it as well.

End result:

.split("\\.")

edited May 14 '14 at 23:23

answered May 14 '14 at 23:17

Jeroen Vannevel

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will fix that , but when i deubg , the datecell array is empty – Exorcismus May 14 '14 at 23:21
@fadytaher And *that* is because the argument to `split` is a regular expression. Escape the `.`. – chrylis -cautiouslyoptimistic- May 14 '14 at 23:22

score 4 · Accepted Answer · edited May 23 '17 at 12:27

split uses regex and in regex . means "any character beside line separators".

So you split on each character creating array full of empty elements like

"foo".split(".")

would at first create ["","","",""], but since split also trails empty elements placed at the end of array you would get empty array []. Trailing last empty strings can be turned off with overloaded version of split split(regex,limit) by passing negative value as limit.

To solve this problem you need to escape .. To do this you can use for instance

split("\\.") - standard escape in regex
split("[.]") - escaping using character class
split("\\Q.\\E") - \Q and \E mark area in which regex metacharacters should be treated as simple literals
split(Pattern.quote(".")) - this method uses Pattern.LITERAL flag inside regex compiler to point that metacharacters used in regex are simple literals without any special meaning

Another problem would be condition in your for loop but more about it in Jeroens answer.

score 1 · Answer 3 · answered May 14 '14 at 23:20

Additionally, you need to escape . when splitting on it, because String.split() treats . as a regex.

public class Foo{
    public static void main(String[] args) {
        String hi = "hello.world.good.bye";
        String[] parts = hi.split("\\.");
        for (int i = 0;i < parts.length; i++)
            System.out.println(parts[i]);
    }
}

String.split on dot doesn't work as expected

3 Answers3