CODEWITHSUNDEEP
javaregex
user2056257
user2056257

Reputation: 141

Regex lookbehind in Java

I am having an issue with lookbehind in java. The following method 

    public static void main (String[] args) throws java.lang.Exception
{
    String num = "1E-12x10";
    String[] numArr = num.split("(?<!E)\\-");
    System.out.println(numArr[0]);

}

produces the output 1E-12x10 as expected - it does not split on the '-'.

The following method 

    public static void main (String[] args) throws java.lang.Exception
{
    String num = "1E-12x10";
    String[] numArr = num.split("[x\\-]");
    System.out.println(numArr[0] + " " + numArr[1] + " " + numArr[2]);

}

also produces the expected output 1E 12 10, splitting on both 'x' and '-'.

But when I try the following method

    public static void main (String[] args) throws java.lang.Exception
{
    String num = "1E-12x10";
    String[] numArr = num.split("[x(?<!E)\\-]");
    System.out.println(numArr[0] + " " + numArr[1] + " " + numArr[2]);

}

I expect the string to split on 'x' but not on '-'. However, what happens is that it splits on 'x', 'E' and '-'. I am not quite sure what's going on here.

Upvotes: 2

Views: 59

Answers (1)

p.s.w.g
p.s.w.g

Reputation: 148990

You can't place lookbehinds inside a character class. You need to use an alternation, like this:

String[] numArr = num.split("x|(?<!E)-");

This will split the string on any x character, or any - not preceded by an E character. Also note that in this case the \\ is not necessary before the -.

Upvotes: 3

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