Regular Expression Help

[Anonymous Monk]

Hi Monks,
I have a string of numbers that comes into my program that supposable, it should have 6 digits long, like 123456, but sometimes it will come in to the program with 4 or 5 digits only, when that happens I have to add 0's to complement the string to have 6 digits.
Like if a string of number comes in with 1234, I have to add 0's to it like, 001234, so it can have 6 digits.
I have a regular expression that isn't working but my question is what is the best way to do this?

if($u=~/\d{4}?/){$u=~s/\d\d\d\d/00$u/;}
if($u=~/\d{5}?/){$u=~s/\d\d\d\d\d/0$u/;}

The code above isn't working because in both cases it match the digits it finds.
Thank you!

[davidrw]

Best way is with sprintf:

$u = sprintf "%06d", $u;

If you want to use a substitution, you could do:

$u =~ s/(\d+)/"0"x(6-length($1)).$1/e; if length($u) < 6;

Update: To correct your attempt:

if($u=~/^\d{4}$/){$u=~s/\d\d\d\d/00$u/;}
if($u=~/^\d{5}$/){$u=~s/\d\d\d\d\d/0$u/;}
# but better written as:
#  $u = "00$u" if $u =~ /^\d{4}$/;
#  $u = "00$u" if length($u) == 4;  # if you know it's digits

[Transient]

$u = ('0' x (6-length($u))).$u;

[Anonymous Monk]

Hi interesting, can you give me an explaination on this code?
Thanks!

[Transient]

Sure! Basically if you take where we started : $u = ('0' x (6-length($u))).$u; we can break this up into multiple statements:

$length_of_u = length($u); # how long is this string?
$how_many_characters_short = 6-$length_of_u;  # how far are we from ha
+ving 6 characters?

$padding = '0' x $how_many_characters_short; # this will give us as ma
+ny 0's as $how_many_characters_short is long (e.g. if it is 4 - it gi
+ves us '0000')

$u = $padding.$u; # concatenate them together... finished!

HTH!

[kirbyk]

There's More Than One Way To Do It!

I prefer: $u = sprintf("%06d", $u);

[Forsaken]

printf is your friend. printf("%06d", $string);(untested) should do the trick.