CODEWITHSUNDEEP
java
Hammelkeule
Hammelkeule

Reputation: 317

Regular Expression Parse Double

I am new to regular expressions. I want to search for NUMBER(19, 4) and the method should return the value(in this case 19,4). But I always get 0 as result !

       int length =0;
       length = patternLength(datatype,"^NUMBER\\((\\d+)\\,\\s*\\)$","NUMBER");

    private static double patternLengthD(String datatype, String patternString, String startsWith) {
        double length=0;
        if (datatype.startsWith(startsWith)) {
            Pattern patternA = Pattern.compile(patternString);
            Matcher matcherA = patternA.matcher(datatype);
                if (matcherA.find()) {
                    length = Double.parseDouble(matcherA.group(1)); 
            }
        }
        return length;
    }

Upvotes: 0

Views: 132

Answers (1)

Andreas
Andreas

Reputation: 159260

You are missing the matching of digits after the comma.
You also don't need to escape the ,. Use this:

"^NUMBER\\((\\d+),\\s*(\\d+)\\)$"

This will give you the first number in group(1) and the second number in group(2).

It is however fairly strict on spaces, so you can be more lenient and match on values like " NUMBER ( 19 , 4 ) " by using this:

"^\\s*NUMBER\\s*\\(\\s*(\\d+)\\s*,\\s*(\\d+)\\s*\\)\\s*$"

In that case you'll have to drop your startsWith and just use the regex directly. Also, you can remove the anchors (^$) if you change find() to matches().

Since NUMBER(19) is usually allowed too. You can make the second value optional:

"\\s*NUMBER\\s*\\(\\s*(\\d+)\\s*(?:,\\s*(\\d+)\\s*)?\\)\\s*"

group(2) will then return null if the second number is not given.

See regex101 for demo.

Note that your code doesn't compile.
Your method returns a double, but length is an int.

Although 19,4 looks like a floating point number, it is not, and representing it as such is wrong.
You should store the two values separately.

Upvotes: 2

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