Java exception pattern

Question

I'm trying to write a code for exception. If the input does not match with the pattern below (this is just example) it will throw an exception message.

8454T3477-90

This is the code that I came up with. However, I'm not sure if this is the right patter...

public void setLegalDescription(String legalDescription) throws MyInvalidLegalDescriptionException
    {
        String valid = ("[0-9999][A-Z][0-9999]-[0-99]");
        if (!legalDescription.matches(valid))
        {
            //throw new MyInvalidLegalDescriptionException("Invalid format! Should be " + "e.g 4050F8335-14");
        }
        this.legalDescription = legalDescription;
    }

use an online tester/debugger for regex to verify regex101.com Tim's answer is correct... — firephil
– firephil, Commented Apr 7, 2019 at 15:32

Unmitigated · Accepted Answer · 2019-04-07 15:54:07Z

6

Your pattern is slightly off. Try this version:

String valid = ("[0-9]{4}[A-Z][0-9]{4}-[0-9]{2}");
if (!legalDescription.matches(valid))
{
    // throw new MyInvalidLegalDescriptionException("Invalid format! Should be " + "e.g 4050F8335-14");
}

An explanation of the regex:

[0-9]{4}   any 4 digits
[A-Z]      any capital letter
[0-9]{4}   any 4 digits
-          a dash
[0-9]{2}   any 2 digits

It should be noted that [0-9999] does not match any number between 0 and 9999. Rather, it actually just matches a single digit between 0 and 9.

If the width of your identifier is not fixed, then perhaps use this pattern:

[0-9]{1,4}[A-Z][0-9]{1,4}-[0-9]{1,2}

edited Apr 7, 2019 at 15:54

Unmitigated

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answered Apr 7, 2019 at 15:29

Tim Biegeleisen

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3 Comments

Jorn Vernee Over a year ago

If the goal is to match any number between 0 and 9999, I think you want {1,4} as a quantifier?

Tim Biegeleisen Over a year ago

@JornVernee I guess that would depend on whether zero would be fixed width (0000), or not (just 0). I am assuming fixed width.

firephil Over a year ago

correct also in the explanation [0-9]{4} to [0-9]{2}

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