CODEWITHSUNDEEP
javaregex
sankethm7
sankethm7

Reputation: 59

suggestion required for Regexp Lookaround in Java

I am using regexp heavily in my project. I need some suggestions for Test strings:

1     string           3.33
      string
      1
      string          -3.33

I need to match the 2nd and 3rd lines (means I do not need string which has 3.33 (currency) at the end of the line). I tried so many variations. The best I got is: 

^[\s]+.+[^(?!(\d+\.\d+))]$

Line 2 matches with this regular expression, but line 3 does not match.

Note: I do care about the beginning or end of the line. So the test lines marked above are with perfect whitespaces.

I use Java as my programming language.

Upvotes: 1

Views: 195

Answers (2)

Alan Moore
Alan Moore

Reputation: 75222

[^(?!(\d+\.\d+))]

is a character class. A character class matches exactly one character from the set of characters you describe within the square brackets. Yours is equivalent to this:

[^!()+.\d]

The ^ at the beginning inverts the set, and \d matches a digit just like it does outside a character class, but the rest of the characters are matched literally. In other words, you're telling it to match any one character that's not !, (, ), +, ., or a digit.

It looks like you were trying to use a negative lookahead, which is a valid approach. If you only care about the dollar amount at the end of the line, you can do this:

^(?!.*\d+\.\d+$).*$

The lookahead tries to match \d+\.\d+ at the end of the line. If it succeeds, the overall match fails. Otherwise, the .*$ consumes the whole line so you can retrieve it with the Matcher's group() method.

This assumes you're applying the regex to one line at a time. If you're trying to find matching lines within a larger text you should specify MULTILINE mode, which you can do like this:

(?m)^(?!.*\d+\.\d+$).*$

Upvotes: 1

Kent
Kent

Reputation: 195039

did you test the text line by line?

then you may use re: \d+\.\d+$ to match the text which you do NOT need. If match() return false, then you take the line.

well it's like grep -v.

if test it with grep:

kent$  cat a
1     string           3.33
      string
      1
      string          -3.33

kent$  grep -Pv '\d+\.\d+$' a
      string
      1

Upvotes: 0

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