regex - Java split is eating my characters

Question

I have a string like this String str = "la$le\$li$lo".

I want to split it to get the following output "la","le\$li","lo". The $ is a $ escaped so it should be left in the output.

But when I do str.split("[^\\]\$") y get "l","le\$l","lo".

From what I get my regex is matching a$ and i$ and removing then. Any idea of how to get my characters back?

Thanks

Answer 1

Use zero-width matching assertions:

    String str = "la$le\$li$lo";
    System.out.println(java.util.Arrays.toString(
        str.split("(?<!\\)\$")
    )); // prints "[la, le$li, lo]"

The regex is essentially

(?<!\)$

It uses negative lookbehind to assert that there is not a preceding .

See also

• regular-expressions.info/Lookarounds

---

More examples of splitting on assertions

Simple sentence splitting, keeping punctuation marks:

    String str = "Really?Wow!This.Is.Awesome!";
    System.out.println(java.util.Arrays.toString(
        str.split("(?<=[.!?])")
    )); // prints "[Really?, Wow!, This., Is., Awesome!]"

Splitting a long string into fixed-length parts, using G

    String str = "012345678901234567890";
    System.out.println(java.util.Arrays.toString(
        str.split("(?<=\G.{4})")
    )); // prints "[0123, 4567, 8901, 2345, 6789, 0]"

Using a lookbehind/lookahead combo:

    String str = "HelloThereHowAreYou";
    System.out.println(java.util.Arrays.toString(
        str.split("(?<=[a-z])(?=[A-Z])")
    )); // prints "[Hello, There, How, Are, You]"

Related questions

• Can you use zero-width matching regex in String split?
• Backreferences in lookbehind
• How do I convert CamelCase into human-readable names in Java?