If you want to do it in one regular expression, without listing each case,
you can use a lookahead assertion:

/^(?=.*8.*)[0-9]{4}$/

The assertion (?=.*8.*) checks that the following matches the expression
contained (.*8.*) which fails if there is not an 8.

2007/2/9, Peter Lauri <[EMAIL PROTECTED]>:

Best group member,



I want to match a four digit number. I allow user to enter with * syntax.
So
8* would match anything that starts with 8 and is 4 digit long so:



/^8[0-9]{3}$/



That was easy. Ok then my other case was: *8, so anything that ends with 8



/^[0-9]{3}8$/



Ok, now the tricky one comes: *8*, so match it incase 8 is anywhere in the
number. Can be beginning, end or in the middle. The problem that I face I
cannot find out a good way of doing this correctly. So I ended up with an
expression like this:



/^(8[0-9]{3}|[0-9]8[0-9]{2}|[0-9]{2}8[0-9]|[0-9]{3}8)$/



This takes care of it and everything, BUT it is so ugly. What I actually
need to construct is: A regular expression that checks if 8 is a part of
the
number, and then that it is four digit long.



The pipe "|" is an OR operator, but are there not any "AND" operator in
Regular Expressions? I have been trying to figure this out for a while
now.
Of course I am using the above syntax right now, but would like to strip
it
down. Maybe not for the performance, but for the beauty of it :-)



If you have any comments and suggestions about this I would be happy.



Best regards,

Peter Lauri



<http://www.dwsasia.com/> www.dwsasia.com - company web site

<http://www.lauri.se/> www.lauri.se - personal web site

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