[SOLVED] Parsing a date

[johng]

I would like to parse out a date in the format (Y-m-d) and get integers for each of them.

Does anyone know how to do this?

[kenrbnsn]

You can use the expode() function:

<?php
$date = '2007-02-28';
$list($year,$month,$day) = explode('-',$date);
?>

 

Ken

 

[Orio]

You can use explode():

 

<?php
//$date holds the date

list($year, $month, $day) = explode("-", $date);

echo "Year: ".$year."<br>Month: ".$month."<br>Day: ".$day;

?>

 

 

Edit- Ken is faster

 

Orio.

[johng]

Thank you both very much!  Do you know if there is any way that I would be able to put that back into the same date format, only adding one day?  I tried using mktime, but it apparently doesn't like variables.  Or if you know of an easier way, that would be awesome!

 

Thanks!

Here's the code I tried to use:

 

 <?php 
list($year, $month, $day) = explode("-", $eserve_date);

echo "Year: ".$year."<br>Month: ".$month."<br>Day: ".$day;
$cst_date = date("Y-m-d", mktime(0,0,0, "$month", "$day", "$year");
?>

 

Before I did the last line, it ran just fine, and had numbers in $month, $day, and $year.

[Orio]

Try this:

 

<?php

$eserve_date = "2007-28-2";
list($year, $month, $day) = explode("-", $eserve_date);

echo date("Y-m-d", strtotime($day."/".$month."/".$year)+(24*60*60));

?>

 

 

Orio.

[johng]

Thank you so much!!!!!  It now works like it is supposed to! 

 

(One minor change that I had to make, I switched $day and $month, but a very minor detail when compared to the help you have given me)

[obsidian]

If you're simply adding one day to a date like that, check out strtotime():

<?php
$date = "2007-02-14";
echo date('Y-m-d', strtotime("$date +1 day"));
?>