How to make a regex syntax in java

Question

I want to make a regex syntax for allowing me ASCII letters, digits, hyphens and underscores and forward slashes.

Currently I am using following syntax but it is not working

if (!id.matches("[a-zA-Z0-9_-/]+")) {
    throw new IllegalArgumentException("Invalid ID: \"" + id
            + "\". An ID may not be the empty string and must contain only ASCII letters, digits, hyphens and underscores and forward slash.");
}

However when I remove the '/' from the regex it is working fine. How can I add forward slash in this condition?

I looked into regex tutorial and it has some examples but those are individual. I want to use forward slash along with ASCII letters, digits.

Escape -. "[a-zA-Z0-9_\-/]+") or move to the end in the character class "[a-zA-Z0-9_/-]+") — Tushar
– Tushar, Commented Aug 1, 2016 at 7:32

Wiktor Stribiżew · Accepted Answer · 2016-08-01 07:34:36Z

5

You create a range with _-/.

Either put the hyphen at the end (and always keep it there):

if (!id.matches("[a-zA-Z0-9_/-]+"))
                            ^

Or, escape the hyphen to always treat it as a literal symbol:

if (!id.matches("[a-zA-Z0-9_\\-/]+"))
                            ^^^

NOTE: There has been some discussion, and it has been pointed out that in case the regular expression is going to be further improved/enhanced/expanded, the best way is to escape the hyphens that should be treated as literal - symbols inside character classes.

edited Aug 1, 2016 at 7:34

answered Aug 1, 2016 at 7:32

Wiktor Stribiżew

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