Regular Expression for Percentage of marks

Question

I am trying to create a regex that matches percentage for marks

For example if we consider few percentages

1)100%
2)56.78%
3)56 78.90%
4)34.6789%
         


        

Answer 1

You haven't double-escaped your dot, which means it's a wildcard for any character, including whitespace.

Use something like:

 ┌ integer part - any 1+ number of digits
 |   ┌ dot and decimal part (grouped)
 |   |┌ double-escaped dot
 |   ||  ┌ decimal part = any 1+ number of digits
 |   ||  |    ┌ 0 or 1 greedy quantifier for whole group
 |   ||  |    |
"\\d+(\\.\\d+)?%"

For instance:

String[] inputs = { "100%", "56.78%", "56 78.90%", "34.6789%" };
Matcher m = null;
for (String s: inputs) {
    m = p.matcher(s);
    if (m.find())
        System.out.printf("Found: %s%n", m.group());
}

Output

Found: 100%
Found: 56.78%
Found: 78.90%
Found: 34.6789%

Note

This still matches the 3rd input, but only the last part.

If you want the 3rd input to just not match, you can surround your pattern with input boundaries, such as ^ for start of input, and $ for end of input.

That would become: "^\\d+(\\.\\d+)?%$"

Or, you can simply invoke Matcher#matches instead of Matcher#find.

Next step

You may want to do something with the numerical value you're retrieving.

In this case, you can surround your pattern with a group ("(\\d+(\\.\\d+)?)%") and invoke either Double.parseDouble or new BigDecimal(...) on your back-reference:

• Double.parseDouble(m.group(1))
• new BigDecimal(m.group(1))