正则表达式（分数的分母）

Question

Soo i think i already solved it, what i did:

String pattern = "(<=|>=)\\s{0,2}((+]\\s{0,2})?(\\d+\\s{0,2}[/]\\s{0,2}(\\d{2,}|[1-9])\\s{0,2}|\\d+[.]\\d{1,2}|\\d+))\\s{0,2}";

The pattern had something wrong, i have corrected it above and now it works :)

I have an inequation that may containing >= or <=, some white spaces and a number. That number might be an integer, a decimal number with 2 decimal places or a fraction and I want to retrieve the number on the 2nd member of the inequation with the "Matcher". Example:

4x1 + 6x2 <= 40/3

I've tried to construct such a pattern and I was able to find it. But then I've remembered that a fraction cannot be divided by zero so I want to check that aswell. For that I have used the following code:

String inequation = "4x1 + 6x2 <= 40/3";
String pattern = "(<=|>=)\\s{0,2}((+]\\s{0,2})?(\\d+\\s{0,2}[/]\\s{0,2}(\\d{2,}|[1-9])\\s{0,2}\\d+|\\d+[.]\\d{1,2}|\\d+))\\s{0,2}";
Pattern ptrn = Pattern.compile(pattern);
Matcher match = ptrn.matcher(inequation);
if(match.find()){
String fraction = match.group(2);
System.out.println(fraction);
} else {
System.out.println("NO MATCH");
}

But it's not working as expected. If it has at least 2 digits on the denominator it returns correctly (eg 40/32). But if it only has 1 digit it only returns the integer part (eg 40).

Anyway to solve this? Which expression should I use?

Answer 1

You could try using debuggex to build regular expressions. It shows you a nice diagram of your expression and you can test your inputs as well.

Java implementation (validates that the numerator is non-zero.):

    Matcher m  = Pattern.compile("[<>]=?\\s{0,2}([0-9]*(/[1-9][0-9]*)?)$").matcher("4x1 + 6x2 <= 40/3");
    if (m.find()) {
        System.out.println(m.group(1));
    }

You need an '$' at the end of your expression, so that it tries to match the entire inequality.

Answer 2

Do you just want the number after the inequality sign? Then do:

Matcher m  = Pattern.compile("[<>]=?\\s*(.+?)\\s*$").matcher(string);
String number = m.find() ? m.group(1) : null;