Adding an AND clause to a regex.

Question

I have this simple regex,

[\d]{1,5}

that matches any integer between 0 and 99999.

How would I modify it so that it didn't match 0, but matches 01 and 10, etc?

I know there is a way to do an OR like so...

[\d]{1,5}|[^0]{1}

(doesn't make much sense)

There a way to do an AND?

Answer 1

My vote is to keep the regex simple and do that as a separate compare outside the regex. If the regex passes, convert it to an int and make sure the converted value is > 0.

But I know that sometimes one regex in a config file or validation property on a control is all you get.

Answer 2

probably better off with something like:

0*[1-9]+[\d]{0,4}

If I'm right that translates to "zero or more zeros followed by at least one of the characters included in '1-9' and then up to 4 trailing decimal characters"

Mike

Answer 3

How about an OR between single digit numbers you will accept and multiple-digit numbers:

^[1-9]$|^\d{2,5}$

Answer 4

I think the simplest way would be:

[1-9]\d{0,4}

throw that between a ^$ if it makes sense in your case, and if so, add a 0* to the beginning:

^0*[1-9]\d{0,4}$

Answer 5

By using look-aheads you can achieve the effect of AND.

^(?=regex1)(?=regex2)(?=regex3).*

Though there is a bug in Internet Explorer, that sometimes doesn't treat (?= ) as zero-width.

http://blog.stevenlevithan.com/archives/regex-lookahead-bug

In your case:

^(?=\d{1,5}$)(?=.*?[1-9]).*

Answer 6

^([1-9][0-9]{0,4}|[0-9]{,1}[1-9][0-9]{,3}|[0-9]{,2}[1-9][0-9]{,2}|[0-9]{,3}[1-9][0-9]|[0-9]{,4}[1-9])$

Not pretty, but it should work. This is more of a brute force approach. There's a better way to do it via grouping as well, but I'm drawing a blank on the actual implementation at the moment.

Answer 7

It looks like you are searching for 2 different conditions. Why not break it out to 2 expressions? It might be simpler and more readable.

var str = user_string;
if ('0' != str && str.matches(/^\d{1,5}$/) {
    // code for match
}

or the following if a string of 0's is not valid as well

var str = user_string;
if (!str.matches(/^0+$/) && str.matches(/^\d{1,5}$/) {
    // code for match
}

Just because you can do it all in one regex doesn't mean that you should.

Answer 8

I think a negative lookahead would work. Try this:

#!/bin/perl -w

while (<>)
{
    chomp;
    print "OK: $_\n" if m/^(?!0+$)\d{1,6}$/;
}

Example trace:

0
00
000
0000
00000
000000
0000001
000001
OK: 000001
101
OK: 101
01
OK: 01
00001
OK: 00001
1000
OK: 1000
101
OK: 101