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Re: Should be a simple spaces/digits regex....but I'm turning grey!
by Bethany (Scribe) on Aug 05, 2014 at 02:11 UTC
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If only digits and space characters are allowed in the string, forget the (numeric) quantifiers and just check for the conditions you described; zero or more spaces followed by one or more digits; anchored fore and aft (at the beginning and end of the string) so nothing can come before the optional spaces and nothing after the digits:
^\s*\d+$
If characters other than digits and spaces may appear, I'd go for a coded solution rather than trying to concoct a single regex to do it all.
(Edited to add "numeric")
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Re: Should be a simple spaces/digits regex....but I'm turning grey! (?=)
by tye (Sage) on Aug 05, 2014 at 02:14 UTC
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# n-1 spaces or digits: vv vvvvv: last digit
(?:(?=\s[\s0-9]|[0-9]{2})[\s0-9]){15}[0-9]
# (^^^^^^^^^^^^^^^^) ^^^^^^^: leading chars
and only allow \s\s, \s\d, or \d\d at each point along the way.
Note that I always use [0-9] and never \d as, these days, \d includes tons of characters besides '0'..'9'.
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Note that I always use 0-9 and never \d as, these days, \d includes tons of characters besides '0'..'9'.
That's news to me. Something to read up on, update rusty skills.
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Yeah, if you've got a version of Perl that supports such. Way too many versions of Perl after \d began including Klingon* digits yet before /a was implemented.
Plus, /a messes with more than just \d. I have yet to run into a single project I was involved in where a string of Klingon* digits would be correctly parsed as a numeric value. But I've touched plenty of projects where \w including more letters than a-z was quite useful. Perl itself is that way, after all. Sure, you could write (?a:\d) but that's just longer and less clear (and less portable).
So I suspect I'll be sticking with [0-9] for quite a while still.
* No, Unicode doesn't actually include Klingon (yet, anyway).
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Re: Should be a simple spaces/digits regex....but I'm turning grey!
by Athanasius (Archbishop) on Aug 05, 2014 at 02:28 UTC
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Hello viffer,
As others have noted, your question is somewhat under-specified. Here is my take on what you may be looking for:
#! perl
use strict;
use warnings;
my $len = 7;
for my $s (' 999', ' 9999', '9999999', ' 9', ' 99 9', '
+9')
{
if (length $s == $len && $s =~ /^\s*\d+$/)
{
printf "%-*s matches\n", $len + 2, "|$s|";
}
else
{
printf "%-*s does not match\n", $len + 2, "|$s|";
}
}
Output:
12:33 >perl 960_SoPW.pl
| 999| matches
| 9999| matches
|9999999| matches
| 9| matches
| 99 9| does not match
| 9| does not match
12:33 >
When the size of the field changes, no need for a new regex — just change the value of $len.
Hope that helps,
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Re: Should be a simple spaces/digits regex....but I'm turning grey!
by Anonymous Monk on Aug 05, 2014 at 02:15 UTC
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' 999' or
' 9999' or
'9999999'
' 9'.
Is "field" length 7 or 8 chars? What is a "field" (part of a larger string)?
but when you're checking a 16 byte filed ZZZZZZZZZZZZ9V99
Um, there is no description of what you want for that one :) one thing at a time?
whilst doing the job,is getting ludicrously large and unreadable
Instead of one regex, write twelve thirteen??
Stop writing unreadable regex :) Write beautiful regex, not ugly, so you can read :) How can I hope to use regular expressions without creating illegible and unmaintainable code?
Someone at work suggested using an sprintf in it's stead within the regex, but I must be honest and say that suggestion has left me clueless on how to do it.
Are you trying to format a string? If you are, go ahead and use sprintf, otherwise ...
write a function?
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ZZZZZZZZZZZZ9V99 is just a COBOL field definition, meaning if the leading digits are 0, don't print them,
hence a value of 9.99 with a definition of ZZ9V99 would print as 9.99,
a definition of 999V99 would result in 009.99 being shown
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Re: Should be a simple spaces/digits regex....but I'm turning grey!
by Laurent_R (Canon) on Aug 05, 2014 at 06:40 UTC
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From your examples and code fragment as much as from your description, it seems that you are looking for a 7-character string made of spaces followed by digits. Why not simply this:
$num = $1 if length($string) == 7 and $string =~ /^\s*(\d+)$/;
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Re: Should be a simple spaces/digits regex....but I'm turning grey!
by BillKSmith (Monsignor) on Aug 05, 2014 at 13:26 UTC
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This passes all your test cases:
use strict;
use warnings;
use Test::Simple qw(no_plan);
my %test_cases = (
' 999' => 'valid',
' 9999' => 'valid',
'999999' => 'valid',
' 9' => 'valid',
' 99 9' => 'invalid',
);
foreach my $case (keys %test_cases) {
my $does_match
= $case =~ /
^\s* # any number of leading spaces
\d+ # followed by a number of digits
(:?
[^\s]* # but cant have any spaces after a digit has b
+een found
\d+ # It must contain at least one digit at the en
+d
)?$
/x ;
ok( !($does_match xor $test_cases{$case} eq 'valid'),
"'$case': $test_cases{$case}" );
}
OUTPUT:
ok 1 - '999999': valid
ok 2 - ' 999': valid
ok 3 - ' 9': valid
ok 4 - ' 9999': valid
ok 5 - ' 99 9': invalid
1..5
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Yes, viffer's solution and mine disagree on the validity of your string. I believe that mine is the one that meets the original specification. In hindsight, it seems that his solution probably does implement what he intendeds.
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Re: Should be a simple spaces/digits regex....but I'm turning grey!
by Anonymous Monk on Aug 05, 2014 at 12:57 UTC
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Tasks like this can also be flummoxed by greed. The greed of a regular expression, that is, not the Deadly Sin. Left to its own devisings, a regex will normally consume the longest substring that qualifies. It must be told not to be so greedy; to take the shortest one instead. Probably not applicable to this case but worth keeping in mind. | [reply] |