FN-FORUM: php date function

date posted 9th May 2006 16:34

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


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