Determining an age

[Joshua Beall]

Hi All,

I am wondering what the best way to determine an age of something/somone is.
For instance, let's say I have two dates:

1985-07-19
2003-11-30

Now, I want to determine the interval in years, and for each fraction of a
year, days. I won't mess with months because they can have variable length,
or if I do measure months I will simply say that a month is 30 days.

So here's my problem: normally when doing date comparison I simply convert
the dates to a UNIX timestamp. This allows me to get the number of seconds
from one date to the other. This is fine for small date ranges. For
instance, let's say I had these dates:

1985-07-19
1985-11-15

Once I get the number of seconds, I simply divide by 86400 in order to
determine the number of days.

But things get complicated once the date range gets large enough, because I
now have to worry about leap years. Every 4 years we have an extra 86400
seconds in a year. So how do I handle this case? If I want to know the
number years and days for two dates that are far apart, like the two I
listed at the beginning, what should I be doing? How do I account for leap
years, where there is an extra day?

I'm looking at PHP's builtin date and time functions, and nothing seems to
handle the case I've got in mind. Do I have to roll my own solution here
and check to see whether or not each year in my daterange is a leap year?

Thanks for any pointers!

Sincerely,
-Josh

[Joshua Beall]

Re: Determining an age

FUT is now set to comp.lang.php, sorry I missed that on the initial posting.



"Joshua Beall" <jbeall@donotsp am.remove.me.he raldic.us> wrote in message
news:pD2Wd.7775 1$wc.33781@trnd dc07...[color=blue]
> Hi All,
>
> I am wondering what the best way to determine an age of something/somone
> is. For instance, let's say I have two dates:
>
> 1985-07-19
> 2003-11-30
>
> Now, I want to determine the interval in years, and for each fraction of a
> year, days. I won't mess with months because they can have variable
> length, or if I do measure months I will simply say that a month is 30
> days.
>
> So here's my problem: normally when doing date comparison I simply convert
> the dates to a UNIX timestamp. This allows me to get the number of
> seconds from one date to the other. This is fine for small date ranges.
> For instance, let's say I had these dates:
>
> 1985-07-19
> 1985-11-15
>
> Once I get the number of seconds, I simply divide by 86400 in order to
> determine the number of days.
>
> But things get complicated once the date range gets large enough, because
> I now have to worry about leap years. Every 4 years we have an extra
> 86400 seconds in a year. So how do I handle this case? If I want to know
> the number years and days for two dates that are far apart, like the two I
> listed at the beginning, what should I be doing? How do I account for
> leap years, where there is an extra day?
>
> I'm looking at PHP's builtin date and time functions, and nothing seems to
> handle the case I've got in mind. Do I have to roll my own solution here
> and check to see whether or not each year in my daterange is a leap year?
>
> Thanks for any pointers!
>
> Sincerely,
> -Josh
>[/color]

[Chung Leong]

Re: Determining an age


"Joshua Beall" <jbeall@donotsp am.remove.me.he raldic.us> wrote in message
news:pD2Wd.7775 1$wc.33781@trnd dc07...[color=blue]
> Hi All,
>
> I am wondering what the best way to determine an age of something/somone[/color]
is.[color=blue]
> For instance, let's say I have two dates:
>
> 1985-07-19
> 2003-11-30
>
> Now, I want to determine the interval in years, and for each fraction of a
> year, days. I won't mess with months because they can have variable[/color]
length,[color=blue]
> or if I do measure months I will simply say that a month is 30 days.
>
> So here's my problem: normally when doing date comparison I simply convert
> the dates to a UNIX timestamp. This allows me to get the number of[/color]
seconds[color=blue]
> from one date to the other. This is fine for small date ranges. For
> instance, let's say I had these dates:
>
> 1985-07-19
> 1985-11-15
>
> Once I get the number of seconds, I simply divide by 86400 in order to
> determine the number of days.
>
> But things get complicated once the date range gets large enough, because[/color]
I[color=blue]
> now have to worry about leap years. Every 4 years we have an extra 86400
> seconds in a year. So how do I handle this case? If I want to know the
> number years and days for two dates that are far apart, like the two I
> listed at the beginning, what should I be doing? How do I account for[/color]
leap[color=blue]
> years, where there is an extra day?
>
> I'm looking at PHP's builtin date and time functions, and nothing seems to
> handle the case I've got in mind. Do I have to roll my own solution here
> and check to see whether or not each year in my daterange is a leap year?
>
> Thanks for any pointers!
>
> Sincerely,
> -Josh
>[/color]

Calculate the way a human would.

First, figure out when was the person's last birthday. If the person was
born in 1985-07-19, then he last celebrated his birthday on 2004-07-19.
Subtract the Unix time of that from the current time and divide by the
number of seconds in a day to get the number of days old.

Subtract the birth year from the current year to get the number of years
old. If the person hasn't celebrated his birthday yet this year, subtract
one.

[Michael Fesser]

Re: Determining an age

.oO(Chung Leong)
[color=blue]
>Calculate the way a human would.
>
>First, figure out when was the person's last birthday. If the person was
>born in 1985-07-19, then he last celebrated his birthday on 2004-07-19.
>Subtract the Unix time of that from the current time and divide by the
>number of seconds in a day to get the number of days old.[/color]

What if the person was born on 1965-07-19?

I would avoid Unix timestamps because of their range restriction and use
some library for date calculations (IIRC there's something in PEAR).
Another way would be to make use of PHP's calendar functions, but these
might not be compiled in by default.

Micha

[Mike]

Re: Determining an age

Hey,

That really sucks huh? And when 2038 (I think) rolls around, the bits will
be maxed out! Amazing how they can create such a solid OS and not even
consider these things.

Anyway, I found a solution at php.net.
Check it out at http://www.php.net/mktime in the comments, or read below!!

Thanks,
Mike

----------------------------------------------------------
If you want to calculate dates before 1970 and your OS doesn't support it
(Windows and Linux), you can easily write your own wrapper to mktime to do
the calculation. The idea is to find a date range in the 1970-2037 time
frame that is the same as your year prior to 1970 (both years start on the
same day of the week and leap years are the same), then add that number of
years to your year value before calling mktime() and then subtract the
corresponding number of seconds from the mktime() result (including the
appropriate number of leap seconds). This way old guys like me (born in
1963) can get a valid number back from mktime() and we can figure out how
old we are in seconds. :-) You can go all the way back to 1902 if you
want.

The mapping is as follows:
1902-1951 = 1986-2035
1952-1969 = 1980-1997
By matching starting day of week for the year and leap years then we should
calculate daylight time correctly (assuming that we have the right rules
set). There's a special hack in the code to turn off daylight time prior to
1942 for date().

Here's the code that we're using now:

//
// Replacement for mktime that does dates before 1970
//
function MakeTime()
{
$objArgs = func_get_args() ;
$nCount = count($objArgs) ;
if ($nCount < 7)
{
$objDate = getdate();
if ($nCount < 1)
$objArgs[] = $objDate["hours"];
if ($nCount < 2)
$objArgs[] = $objDate["minutes"];
if ($nCount < 3)
$objArgs[] = $objDate["seconds"];
if ($nCount < 4)
$objArgs[] = $objDate["mon"];
if ($nCount < 5)
$objArgs[] = $objDate["mday"];
if ($nCount < 6)
$objArgs[] = $objDate["year"];
if ($nCount < 7)
$objArgs[] = -1;
}
$nYear = $objArgs[5];
$nOffset = 0;
if ($nYear < 1970)
{
if ($nYear < 1902)
return 0;
else if ($nYear < 1952)
{
$nOffset = -2650838400;
$objArgs[5] += 84;
// Apparently dates before 1942 were never DST
if ($nYear < 1942)
$objArgs[6] = 0;
}
else
{
$nOffset = -883612800;
$objArgs[5] += 28;
}
}

return call_user_func_ array("mktime", $objArgs) + $nOffset;
}

In Linux, the values returned will work just fine with date() and I believe
that strftime() should work too, but Windows doesn't like the negative
values at all. We had to write a replacement for strtotime since it appears
to call mktime() to come up with the final result.

I hope that this helps someone.

- todd



"Joshua Beall" <jbeall@donotsp am.remove.me.he raldic.us> wrote in message
news:pD2Wd.7775 1$wc.33781@trnd dc07...[color=blue]
> Hi All,
>
> I am wondering what the best way to determine an age of something/somone
> is. For instance, let's say I have two dates:
>
> 1985-07-19
> 2003-11-30
>
> Now, I want to determine the interval in years, and for each fraction of a
> year, days. I won't mess with months because they can have variable
> length, or if I do measure months I will simply say that a month is 30
> days.
>
> So here's my problem: normally when doing date comparison I simply convert
> the dates to a UNIX timestamp. This allows me to get the number of
> seconds from one date to the other. This is fine for small date ranges.
> For instance, let's say I had these dates:
>
> 1985-07-19
> 1985-11-15
>
> Once I get the number of seconds, I simply divide by 86400 in order to
> determine the number of days.
>
> But things get complicated once the date range gets large enough, because
> I now have to worry about leap years. Every 4 years we have an extra
> 86400 seconds in a year. So how do I handle this case? If I want to know
> the number years and days for two dates that are far apart, like the two I
> listed at the beginning, what should I be doing? How do I account for
> leap years, where there is an extra day?
>
> I'm looking at PHP's builtin date and time functions, and nothing seems to
> handle the case I've got in mind. Do I have to roll my own solution here
> and check to see whether or not each year in my daterange is a leap year?
>
> Thanks for any pointers!
>
> Sincerely,
> -Josh
>[/color]

[Chung Leong]

Re: Determining an age

"Michael Fesser" <netizen@gmx.ne t> wrote in message
news:a35i21l1ie pfjjpp680h1qjg7 mdoudgh8r@4ax.c om...[color=blue]
> .oO(Chung Leong)
>[color=green]
> >Calculate the way a human would.
> >
> >First, figure out when was the person's last birthday. If the person was
> >born in 1985-07-19, then he last celebrated his birthday on 2004-07-19.
> >Subtract the Unix time of that from the current time and divide by the
> >number of seconds in a day to get the number of days old.[/color]
>
> What if the person was born on 1965-07-19?[/color]

His birthday would still be July 19th. That doesn't change anything.

[Michael Fesser]

Re: Determining an age

.oO(Chung Leong)
[color=blue]
>"Michael Fesser" <netizen@gmx.ne t> wrote
>[color=green]
>> What if the person was born on 1965-07-19?[/color]
>
>His birthday would still be July 19th. That doesn't change anything.[/color]

The age calculation using Unix timestamps will fail.

Micha

[Chung Leong]

Re: Determining an age


"Michael Fesser" <netizen@gmx.ne t> wrote in message
news:vhnm21duls gq2vnhtc79gjk6g d9890nt8k@4ax.c om...[color=blue]
> .oO(Chung Leong)
>[color=green]
> >"Michael Fesser" <netizen@gmx.ne t> wrote
> >[color=darkred]
> >> What if the person was born on 1965-07-19?[/color]
> >
> >His birthday would still be July 19th. That doesn't change anything.[/color]
>
> The age calculation using Unix timestamps will fail.
>[/color]

Not for another three decades. Read the original message and my response
again.

[Michael Fesser]

Re: Determining an age

.oO(Chung Leong)
[color=blue]
>Not for another three decades. Read the original message and my response
>again.[/color]

Got it. Sorry, my fault. Read too fast.

Micha