regular expression matching all 8 character strings except "00000000" - regex

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

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.

Answer 4

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 5

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 6

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 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

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?