regular expression matching all 8 character strings except "00000000"

Question

I am trying to figure out a regular expression which matches any string with 8 symbols, which doesn't equal "00000000".

can any one help me?

thanks

Answer 1

In at least perl regexp using a negative lookahead assertion: ^(?!0{8}).{8}$, but personally i'd rather write it like so:

length $_ == 8 and $_ ne '00000000'

Also note that if you do use the regexp, depending on the language you might need a flag to make the dot match newlines as well, if you want that. In perl, that's the /s flag, for "single-line mode".

Answer 2

Unless you are being forced into it for some reason, this is not a regex problem. Just use len(s) == 8 && s != "00000000" or whatever your language uses to compare strings and lengths.

Answer 3

Unless you have unspecified requirements, you really don't need a regular expression for this:

if len(myString) == 8 and myString != "00000000":
    ...

(in the language of your choice, of course!)

Answer 4

If you need to extract all eight character strings not equal to "000000000" from a larger string, you could use

"(?=.{8})(?!0{8})."

to identify the first character of each sequence and extract eight characters starting with its index.

Answer 5

Wouldn't ([1-9]*|\D){8} do it? Or am I missing something here (which is actually just the inverse of ndim's, which seems like it oughta work).

I am assuming the characters was chosen to include more than digits.

Ok so that was wrong, so Professor Bolo did I get a passing grade? (I love reg expressions so I am really curious).

>>> if re.match(r"(?:[^0]{8}?|[^0]{7}?|[^0]{6}?|[^0]{5}?|[^0]{4}?|[^0]{3}?|[^0]2}?|[^0]{1}?)", '00000000'):
    print 'match'
... 
>>> if re.match(r"(?:[^0]{8}?|[^0]{7}?|[^0]{6}?|[^0]{5}?|[^0]{4}?|[^0]{3}?|[^0]{2}?|[^0]{1}?)", '10000000'):
...     print 'match'
match
>>> if re.match(r"(?:[^0]{8}?|[^0]{7}?|[^0]{6}?|[^0]{5}?|[^0]{4}?|[^0]{3}?|[^0]{2}?|[^0]{1}?)", '10011100'):
...     print 'match'
match
>>> 

That work?

Answer 6

As mentioned in the other answers, regular expressions are not the right tool for this task. I suspect it is a homework, thus I'll only hint a solution, instead of stating it explicitly.

The regexp "any 8 symbols except 00000000" may be broken down as a sum of eight regexps in the form "8 symbols with non-zero symbol on the i-th position". Try to write down such an expression and then combine them into one using alternative ("|").

Answer 7

Of course, one would simply check

if stuff != '00000000'
   ...

but for the record, one could easily employ heavyweight regex (in Perl) for that ;-)

...
use re 'eval';

my @strings = qw'00000000 00A00000 10000000 000000001 010000';
my $L = 8;

print map "$_ - ok\n",
      grep /^(.{$L})$(??{$^Nne'0'x$L?'':'^$'})/,
      @strings;

...

prints

00A00000 - ok
10000000 - ok

go figure ;-)

Regards

rbo

Answer 8

If you need a regex, ^(?!0{8})[A-Za-z0-9]{8}$ will match a string of exactly 8 characters. Changing the values inside the [] will allow you to set the accepted characters.