How to get all integers before hyphen from java String

Question

I want to parse through hyphen, the answer should be 0 0 1 (integer), what could be the best way to parse in java

public static String  str ="[0-S1|0-S2|1-S3, 1-S1|0-S2|0-S3, 0-S1|1-S2|0-S3]";

Please help me out.

Answer 1

Use the below regex with Pattern and matcher classes.

Pattern.compile("\\d+(?=-)");

\\\\d+ - Matches one or more digits. + repeats the previous token \\\\d ( which matches a digit character ) one or more times.

(?=-) - Only if it's followed by an hyphen. (?=-) Called positive lookahead assertion which asserts that the match must be followed by an - symbol.

String str ="[0-S1|0-S2|1-S3, 1-S1|0-S2|0-S3, 0-S1|1-S2|0-S3]";
Matcher m = Pattern.compile("\\d+(?=-)").matcher(str);
while(m.find())
{
    System.out.println(m.group());
}

Answer 2

one lazy way: if you already know the pattern of the string, use substring and indexof to locate your word.

String str ="[0-S1|0-S2|1-S3, 1-S1|0-S2|0-S3, 0-S1|1-S2|0-S3]";
integer int1 = Integer.parseInt(str.substring(str.indexOf("["),str.indexOf("-S1")));

and so on.