CODEWITHSUNDEEP
javaregexstring
Anusha
Anusha

Reputation: 969

Splitting characters using regex returns an empty value in java

Here is my input

....ALPO..LAPOL.STRING

I want to seperate each string when it reaches '.' and store in string array list.

I tried using the below code,

  ArrayList<String> words = new ArrayList<>    (Arrays.asList(chars.stream().map(String::valueOf)
      .collect(Collectors.joining("")).split("\\.+")));

There is a problem with regex split("\.+")) .

EXPECTED OUTPUT:

ALPO

LAPOL

STRING

ACTUAL OUTPUT:

" " - > Blank Space

LAPOL

STRING

It prints the first value of the list as an empty value because there are many '.' present before 'A". How to get rid of this empty value in string array list. Any help would be glad !!

Upvotes: 1

Views: 558

Answers (1)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626758

The empty element appears because the delimiters are matched before the first value you need to get.

You need to remove the delimiter symbols from the start of the string first with .replaceFirst("^\\.+", "") and then split:

String results[] = "....ALPO..LAPOL.STRING".replaceFirst("^\\.+", "").split("\\.+");
System.out.println(Arrays.toString(results));

See the IDEONE demo

The ^\\.+ pattern matches the beginning of a string (^) and then 1 or more literal dots (\\.+). replaceFirst is used because only 1 replacement is expected (no need using replaceAll).

A bit more details on splitting in Java can be found in the documentation:

public String[] split(String regex)
Splits this string around matches of the given regular expression. This method works as if by invoking the two-argument split method with the given expression and a limit argument of zero. Trailing empty strings are therefore not included in the resulting array.

However, leading empty elements will be included if found. So, we first need to get rid of those delimiters at the string start.

Upvotes: 5

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