Date math problem

[tony_l]

I need to know how you can subtract two dates.

I have a MySQL Database full of profile information. I need the table to display an image that depends on how long the person has been a member.

The problem i am having is that it is not having the result i thought it would have and it is because my date math is not working correctly.

Can you look at the following code and tell me-

1. How to do the correct date math?

2. What type of an expression do I use so that the correct award is shown (i.e. members that have been inservice BETWEEN 1 month and 2 months)?

#######################

Get the time info

#######################

$format = getdate(time());
$year = $format['year'];
$month = $format['mon'];
$day = $format['mday'];

$checkdate = "$year-$month-$day";

######################################

Get ready for the Display Loop!!

######################################

while ($row1 = mysql_fetch_array($Result1)){

IF ($row1[datejoin]-$checkdate >= 0000-01-00) {

$award = "<img src='http://www.mysite.com/images/Awards/medal2.gif'>";

}

IF ($row1[datejoin]-$checkdate >= 0000-02-00) {

$award = "<img src='http://www.mysite.com/images/Awards/medal0.gif'>";

}

print("<TR VALIGN=LEFT>");

print("<TD VALIGN=LEFT>$award</TD>");

print("<TD VALIGN=LEFT>$row1[user]</TD>");

print("<TD VALIGN=LEFT>$row1[position]</TD>");

print("<TD VALIGN=LEFT>$row1[datejoin]</TD>");

print("</TR>");
}

[Anon]

The easiest is to take both dates, convert them to Unix timestamps, and take the difference of those.

That will give you the difference in seconds; then (int)($difference/(246060)) will be the difference in days.

$weeks=(int)($difference/(2460607));
$difference%=2460607;
$days=(int)($difference/(246060));