RE: FN-FORUM: php date function

date posted 9th May 2006 16:49

You have a few ways of doing this

First is to get the mysql date yyyy-mm-dd and do an explode() on it

$date_array = explode('-','2006-01-01')

Will give you an array of

$date_array = array(
0 => 2006,
1 => 01,
2 => 01
);


Or look up the mysql date functions

Ash

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-----Original Message-----
From: [EMAIL REMOVED] [EMAIL REMOVED] On Behalf Of Mark Bell
Sent: 09 May 2006 17:33
To: FN-FORUM / [EMAIL REMOVED]
Subject: FN-FORUM: php date function


Afternoon brain teaser:

I'm storing dates in a MySQL table (of last client visits) and want a count
of how many days since the last visit.

I'm storing the date in an 2006-01-01 format.

I've grabbed this: (from http://us3.php.net/date)

$date1 = mktime(0,0,0,1,1,2006);
$date2 = mktime(0,0,0,date('m'),date('d'),date('Y'));
$total_days = 0;
while($date1 < $date2) { $total_days++; $date1 += 86400; }
echo $total_days;

but I guess is need to use something like this:

$date1 = mktime(0,0,0,$lastinmonth,$lastinday,$lastinyear);
$date2 = mktime(0,0,0,date('m'),date('d'),date('Y'));
$total_days = 0;
while($date1 < $date2) { $total_days++; $date1 += 86400; }
echo $total_days;

so how do I convert the 2006-01-01 record in my dB into the 3 required
variables as above? ($lastinmonth,$lastinday,$lastinyear).

or maybe there is another way of outputting it?

I simply want to say, you where in $givendate, today is $todaysdate which is
$dayspast since your last visit.

Thoughts welcome!
Thanks


---
Mark Bell

mob: 0774 086 1862
office: 0870 383 4575
fax: 0870 199 2149
skype: m4rkbell, 020 8123 2862
web: http://m4rk.biz
email: [EMAIL REMOVED]




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