CODEWITHSUNDEEP
javaregex
Peter Eriksson
Peter Eriksson

Reputation: 239

Problem with % and / in Java regex

I have a problem with the % and / characters in Java regex. The following example will illustrate my issue:

Pattern pattern = Pattern.compile("^[a-z]*[/%]$");
Matcher m = pattern.matcher("a%/");
System.out.println(m.find());

It prints "false" when I expect it to be "true". The % and / sign shouldn't have to be escaped but even if I do it still dosn't work.

So my question is simply why?

Upvotes: 1

Views: 97

Answers (3)

F.P
F.P

Reputation: 17831

You just check for one of the [%/] characters to appear, not the possibility of having both.

Try [%/]? (or +, *, depending what you want)

Upvotes: 0

Amarghosh
Amarghosh

Reputation: 59451

^[a-z]*[/%]$ matches zero or more lower case letters followed by one character which can be either / or % - to allow multiple characters, use

^[a-z]*[/%]+$

+ stands for one or more; use * for zero or more.

If you didn't have $ at the end of the regex, it would have matched a% in the stringa%/.

$ matches end of line.

Upvotes: 5

Quentin
Quentin

Reputation: 943981

Your regular expression says "zero or more lower case letters then / or % then the end of the string".

The string matches until it gets a / when it was looking for the end of the string.

You probably want to remove the square brackets to say "/ then %"

Upvotes: 1

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