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Re: [PHP] Re: Date()

Posted by Steve Buehler on 01/17/05 23:36

At 08:33 PM 1/15/2005, you wrote:

>Torsten,
>Whatever the combination, it echos "February 02-2005<br>February
>02-2005<br>February 02-2005". What is wrong with it?
>
> <?php
> $week5 = "2005-02-14";
> $firstDayTs = strtotime($week5);
> $lastDayTs = $firstDayTs + (4 * 86400);
> echo date('F', $firstDayTs) . ' ' . date('m', $firstDayTs) . '-' .
> date('Y',$firstDayTs) ."<br>";
> echo date('F', $firstDayTs) . ' ' . date('m', $firstDayTs) . '-' .
> date('Y',$lastDayTs) ."<br>";
> ?>
>
>John

It is outputing what you are asking it to:
F A full textual representation of a month, such as January or March
m Numeric representation of a month, with leading zeros
Y A full numeric representation of a year, 4 digits
This is all documented at: http://us2.php.net/manual/en/function.date.php

First you are asking for the Month as a name, then the month as a number
then the year as a number. If you made $week5=2005-12-31 it would give you
the following output:
December 12-2005<br>December 12-2006<br>
If you are wanting to make it echo "February 14-18", then check out the 3rd
echo statement in the below example.

<?php
$week5 = "2005-02-14";
$firstDayTs = strtotime($week5);
$lastDayTs = $firstDayTs + (4 * 86400);
echo date('F', $firstDayTs) . ' ' . date('d', $firstDayTs) . '-' .
date('Y',$firstDayTs) ."<br>\n";
echo date('F', $lastDayTs) . ' ' . date('d', $lastDayTs) . '-' .
date('Y',$lastDayTs) ."<br>\n";
echo date('F', $lastDayTs) . ' ' . date('d', $firstDayTs) . '-' .
date('d',$lastDayTs) ."<br>\n";
echo "\n";
?>
Now, the problem with that is that if your $week5="2005-02-18" then your
results will be wrong....unless this is what you want, but the answer will
be "February 28-04". You are outputing the Month from $week5, the day from
$week5 and the day+4 of $week5 which takes it into the next month.

Steve

 

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