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
<http://www.carbonfree.org.uk/> www.carbonfree.org.uk - become Carbon Free
