Regular Expressions

[Vaibhav G Garg]

Hi All,

We need to write a regular expression which can accept a list of alphanumeric characters separated by single commas. The list can't start with the , but can end with a ,


Length of the characters will be from 1 to 9. So, considering following examples and their outcomes:

123,45a93, - PASS
1234567890,1939 - FAIL - Length is more than 10 chars for 1st one
123,,,393 - FAIL - More than one ,
123,455 - PASS

I have written the following regular expression:

(^(([1-9a-zA-Z]{1,9}),)+$);

Here, it is working fine for all the cases EXCEPT ONE i.e. it always expects the , in the end of the string. But, as per my requirement the string can end with a , or may not end with a , as well. I tried putting ,? but then it starts accepting more than 9 characters since it accepts 0 or 1 , between two entries.

Please suggest how to resolve this issue. Thanks in advance.

Thanks,
Vaibhav Garg

[Jeanne Boyarsky]

Do you have to use a regular expression for the whole thing? It would be easier to write if (s.charAt(0) != ',' && s.matches(yourRegex) rather than complicating the regular expression.

[Richard Tookey]

You just need to replace the ',' in the regular expression so that it accepts as terminator a comma OR end of line ie '(,|$)' .

[Vaibhav G Garg]

I am able to write down the regular expression for this:

exp = "[1-9a-zA-Z]{1,9}";
regex = new RegExp("^(" + exp + ",)+$|^" + exp + "(," + exp + ")*$");

Thanks all for your inputs.

[Richard Tookey]

That is much more complicated than the approach I suggested!

[Vaibhav G Garg]

Yes, I agree Richard Tookey. Your approach is much more simpler. Thanks a lot for your inputs.