[SOLVED] Date produce

[timmah1]

Hi

 

How would I take this code

$start = date("Y-m-d");
$formatted_date = date("M j, Y", strtotime($start));

And make it echo 3 days ahead?

 

Meaning, today's date is Jan. 15, 2008,  I would like this code to be able to distinguish 3 days ahead, meaning it would echo out Jan. 18, 2008.

 

I've tried this, but it only shows the 3

$start = date("Y-m-d");
$formatted_date = date("M j, Y", strtotime($start));
$ends = $formatted_date+3;

 

Thanks in advance

[cooldude832]

<?php
$start = date("Y-m-d");
$formatted_date = date("M j, Y", strtotime("+3 Days", $start));
echo "Start: ".$start."<br />Formatted: ".$formatted_date;
?>

[timmah1]

With that, it displays this

	Start: 2008-01-15
Formatted: Jan 3, 1970

[cooldude832]

your version of php doesn't support a start date in that format so try (the yyyy-mm-dd)

so try

<?php
$start = date("U");
$formatted_date = date("M j, Y", strtotime("+3 Days", $start));
echo "Start: ".$start."<br />Formatted: ".$formatted_date;
?>

 

really if you are strating at NOW() you can ignore the start and just do

<?php
$formatted_date = date("M j, Y", strtotime("+3 Days"));
echo "Formatted: ".$formatted_date;
?>

[timmah1]

Works perfect.

 

Thank you very much.

 

What version is this php then? hmmm

 

Thank you

[cooldude832]

strtotime is on the flaw list of php for not being able to do things consistently which is a problem cause it doesn't know yyyy-mm-dd from yyyy-dd-mm from mm-dd-yyyy from dd-mm-yyyy

 

 

[timmah1]

Thank you, you are very much a 'super guru'

 

I really appreciate your help

[PFMaBiSmAd]

The problem with the above code is that the second parameter of strtotime() is a UNIX Timestamp.

 

The original code inserted a formatted string instead and produces a Notice warning and outputs a 1970 date.