Date using php

[musicbase]

Hi there,

I have been looking through this forum and discovered that the way most people add X number of days to today's date is using this code:

echo mktime(0, 0, 0, date("n"), date("j") + 10, date("Y")); 

When I do this, I get an output similar to this - 1156114800

I am not sure what I can do with this, I am trying to compare it to a date pre-entered into mysql. I also tried using strtotime("+1 day"), with a similar output again.

Thanks in advance,
Richard Stephens

[ednark]

you are having a formatting problem, you must UTSL and RTFM and type things into google all that good stuff about the php time functions and the mysql time functions in order so recify this.

mysql native date format as a string that must be parsed into useful numbers

mktime is a php function that returns a unix time stamp, an integer

you have to either convert mysql time to a like integer and compare them with UNIX_TIMESTAMP sql function

or find a php function to turn a unix timestamp into a sql formatted string and compare them.

[Hans_Verstraele]

As an example I added 2 hours to the current timestamp and then formatted output for the month, day and year. Read the php manual for the date() function to see further explanation. Hope that helps.

$unix_first_delivery = mktime (date("H")+2 ,0 ,0 ,date("m") ,date("d")+2 ,date("Y"));

$first_delivery_month = (date("n",$unix_first_delivery)); // month nr (1..31)
$first_delivery_day = (date("j",$unix_first_delivery)); // day nr (0..6)
$first_delivery_year = (date("Y",$unix_first_delivery)); // year nr (2004..)

echo "month = ".$first_delivery_month."<br>";
echo "day = ".$first_delivery_day."<br>";
echo "year = ".$first_delivery_year."<br>";

/Hans