On Jul 7, 2005, at 7:54 PM, Bruno B B Magalhães wrote:
Hi Folks,
Well I think I got it, at least it's working more or less to me now...
I really would appreciate any comments or suggestions.
function str2time($input = '12/31/1969')
{
if(($output = strtotime($input)) !== -1)
{
return $output;
}
else
{
preg_match('([0-2][0-9][0-9][0-9])', $input, $year);
preg_replace('([0-2][0-9][0-9][0-9])', '1976', $input);
return floor(strtotime($input) + (($year[0] - 1976) *
(31557376.189582)));
}
}
Thanks a lot!
Best Regards,
Bruno B B Magalhaes
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One problem is that there's no accounting for leap years, but I don't
know if that's gonna cause you any problems or not.
The other thing I noticed is that the 'match' bit you have in there as
"year[0]" should be "$year[1]". $year[0] will return the
whole string that was matched, not just the actual match part between
the parenthesis. Although I think you would get the
same thing with your set up. And you will need to put a delimiter in
the regex part, right now it looks like it's going to treat
the parenthesis as the delimiter which will make the return for the
match not work. ie:
" preg_match('([0-2][0-9][0-9][0-9])', $input, $year); " -- $year will
be empty...
should be " preg_match('/([0-2][0-9][0-9][0-9])/', $input, $year); "
or " preg_match('/([0-1][0-9][0-7][0-9])/', $input, $year); " -- to
restrict it to 1970 or before....
but it could also be " preg_match('/([\d]{4})/', $input, $year); " --
if you don't really need to validate the the year
There's other problems that I can see with the math logic in the
return, like why 1976?, why would you want to generate
a positive number that will conflict with dates before 1970? but it
could just be that I'm not thinking the math all the way
through, and what you eventually want to do with the dates once you
store them.
Edward Vermillion
evermillion@xxxxxxxxxxxx
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