Re: Y2K Again ??

$Yr = ($FieldA =~ /^(\d{2}).*/ and $1 > 20 ? 19 : 20).$1;

On May 14, 2004, at 8:25 AM, Rick Klement wrote:

> Brian Morgan wrote:
>>
>> Good evening everybody,
>> I have been working on a small database project and have come across
>> the
>> need to move some data from table A to table B, the year format in
>> table
>> A is \d{2}\D{8} and in table B \d{4}\D{8}
>>
>> Since the records in table A range from 1970 - 2004.  I want to be
>> able
>> look at the date, and determine if it is a 2000 or 1900 date and make
>> the appropiate date change.
>>
>> So here is my sollution (long):
>> #Pull the value out of the field
>> #For this example I don't care about the rest of FieldA I just want
>> the
>> 1st 2 digits,
>> #but I need to preserve FieldA for later use in the program.
>>
>> ($Yr = $FieldA) =~ s/^(\d{2}).*/$1/;
>> if($Yr > 20){
>>     $Yr = "19$Yr";
>> }else{
>>     $Yr = "20$Yr";
>> }
>>
>> #Concat year to FieldB
>>
>> I guess I am looking for a way to combine a substitution with the if
>> statement but haven't come up with a working solution.
>>
>> Anyways, I am looking forward to see what the outcome is.
>>
>> Brian
>
> There is no need for a regex
>
> 	$Yr = ($FieldA > 20 ? 1900 : 2000) + $FieldA;
>
> Or just get $FieldB directly if you don't need a separate $Yr
>
> 	$FieldB = ($FieldA < 21) + 19 . $FieldA;
>
> --
> Rick Klement
>