正则表达式解析双精度

Question

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;
    }

Answer 1

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.