Date problems

[wispa]

I'm having trouble with some date functionality which I'm hoping someone will be able to help me with.

The scenario is that I'm being given data which has a creation date. The problem is that the creation date is calculated by the number of hours passed from a pre-defined date, in this case 1900-01-01.

I've tried a couple of methods to work this out, my first looks like this:

[code=php]
$datefrom = 957047;
$this->strDate = (date('Y-m-d H:00:00', strtotime('+'.$datefrom.' hours', date_format(date_create('1900-01-01'), 'U')))."\n");
[/code]

This ends up printing out '2011-02-16 19:00:00'
The date I should be getting should be something in the region of '2009-01-02'
This method is basically trying to use strtotime to increase the given date by the given number of hours.

My second method looks like this:

[code=php]
$datefrom = 957047;
$secondsSince1900 = date_format(date_create('1900-01-01'), 'U');
$datefromInSeconds = $datefrom*60*60;

$SecondsAdded = $secondsSince1900+$datefromInSeconds;

$newDate = date_create($SecondsAdded);

echo("dateformat: ".date('Y-m-d H:00:00', $SecondsAdded)."\n");
[/code]

This one's getting a bit closer with a result of '2009-03-06 23:00:00'
I've converted the hours passed into seconds and added that to the timestamp, then I've converted it into a date format.

I did notice that if I change the '$datefrom' to 0, then the result is actually '1901-12-13 20:00:00' instead of '1900-01-01' which it should still be as it's just changing the date from a timestamp and back to a date format.

Any help would be fantastic.

Thanks in advance.

[AnthonySterling]

If you're just adding hours to a time stamp, could you not just use the following?

<?php
function add_hours_to_date($fHours, $iTimestamp)
{
    return  $iTimestamp + (($fHours * 60) * 60);
}

$iNow = time();
echo date('r', $iNow); #Mon, 11 May 2009 09:52:14 +0100

$iNow = add_hours_to_date(3.5, $iNow);
echo date('r', $iNow); #Mon, 11 May 2009 13:22:14 +0100
?>

[Dan Grossman]

SilverBulletUK: The way I read it, he's being given a date in the form of "hours since 1900-01-01", which is not the same as adding hours to a timestamp.

The problem is that 1900-01-01 is before the Unix epoch, so you can't use date_format('U') to turn it into a timestamp to add hours to. It's before time 0.

So you need to do some math yourself before you can work with the number as a timestamp.

Start by figuring out how far from 1900-01-01 time 0 (1970-01-01) is, and subtracting that from your $datefrom

        $datefrom = 957047;

        $base = 70 * 365.25 * 24 * 60 * 60;
        $timestamp = ($datefrom * 60 * 60) - $base;

        echo date('Y-m-d', $timestamp); 

Now you have a Unix timestamp for the creation date (2009-03-06).

Does that sound right?

I'm not sure what to do about the leap years though. Multiplying by 365.25 is a hack that could leave you a day off..

[wispa]

Thanks for the reply, that looks like it's essentially what my second block of code is doing, just inline instead of a function.

$datefrom = 957047;
$secondsSince1900 = date_format(date_create('1900-01-01'), 'U');
$datefromInSeconds = $datefrom*60*60;
$SecondsAdded = $secondsSince1900+$datefromInSeconds;
$newDate = date_create($SecondsAdded);
echo("dateformat: ".date('Y-m-d H:00:00', $SecondsAdded)."\n"); 

I don't think this is giving the correct result though.... it's returning '2009-03-06 23:00:00' when I've been led to believe that the result should be '2009-01-02'...

I shall have to try this method with some other data to see what the results are like.

Thanks again.

[wispa]

SilverBulletUK: The way I read it, he's being given a date in the form of "hours since 1900-01-01", which is not the same as adding hours to a timestamp.

The problem is that 1900-01-01 is before the Unix epoch, so you can't use date_format('U') to turn it into a timestamp to add hours to. It's before time 0.

So you need to do some math yourself before you can work with the number as a timestamp.

Start by figuring out how far from 1900-01-01 time 0 (1970-01-01) is, and subtracting that from your $datefrom

        $datefrom = 957047;

        $base = 70 * 365.25 * 24 * 60 * 60;
        $timestamp = ($datefrom * 60 * 60) - $base;

        echo date('Y-m-d', $timestamp); 

Now you have a Unix timestamp for the creation date (2009-03-06).

Does that sound right?

I'm not sure what to do about the leap years though. Multiplying by 365.25 is a hack that could leave you a day off..

Well that's something.... it's a different method but gives the same result to what I was getting....... it's possible I was given an incorrect result to compare it to.

I shall look into this

Thanks

[Dan Grossman]

It can't be 2009-01-02...

957047 / 24 = 39876.96 days
39875.96 / 365 = 109.25 years

A quarter of a year would put you in March, not January

[AnthonySterling]

What about the DateTime object? Is this viable?

<?php
$oDate = new DateTime('01-01-1900', new DateTimeZone('Europe/London'));

$oDate->modify(
    sprintf(
        '+ %s Hours',
        278324
    )
);

echo $oDate->format('r'); #Fri, 02 Oct 1931 20:00:00 +0100
?>

[wispa]

It can't be 2009-01-02...

957047 / 24 = 39876.96 days
39875.96 / 365 = 109.25 years

A quarter of a year would put you in March, not January

A very good point, I've asked the data provider on their thoughts on this.

[wispa]

It can't be 2009-01-02...

957047 / 24 = 39876.96 days
39875.96 / 365 = 109.25 years

A quarter of a year would put you in March, not January

I've received word that the date within some of the files is incorrect. The date is also encoded within the filename which is correct and safe to use.

Some people!

It's a good thing the file I was testing contained faulty data so it could be caught at this stage of the project!

Thanks for all the help.