How to split String with this regular expression?

Question

if (url.contains("|##|")) {
    Log.e("url data", "" + url);
    final String s[] = url.split("\\|##|");
}

I have a URL with the separator "|##|"

I tried to separate this but didn't find solution.

Answer 1

Use Pattern.quote, it'll do the work for you:

Returns a literal pattern String for the specified String.

final String s[] = url.split(Pattern.quote("|##|"));

Now "|##|" is treated as the string literal "|##|" and not the regex "|##|". The problem is that you're not escaping the second pipe, it has a special meaning in regex.

An alternative solution (as suggested by @kocko), is escaping^* the special characters manually:

final String s[] = url.split("\\|##\\|");

^{* Escaping a special character is done by \, but in Java \ is represented as \\}

Answer 2

You have to escape the second |, as it is a regex operator:

final String s[] = url.split("\\|##\\|");