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.
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3 Answers
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 \\
2 Comments
Konstantin Yovkov
I like this approach (
Pattern.quote)
Rahul Tripathi
That was super quick! +1
You have to escape the second |, as it is a regex operator:
final String s[] = url.split("\\|##\\|");
Comments
You should try to understand the concept as well - String.split(String regex) interprets the parameter as a regular expression, and since pipe character "|" is a logical OR in regular expression, you would be getting result as an array of each alphabet is your word.
Even if you had used url.split("|"); you would have got same result.
Now why the String.contains(CharSequence s) passed the |##| in the start because it interprets the parameter as CharSequence and not a regular expression.
Bottom line: Check the API that how the particular method interprets the passed input. Like we have seen, in case of split() it interprets as regular expression while in case of contains() it interprets as character sequence.
You can check the regular expression constructs over here - http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html
