[snip]
I need to subtract months by their abbreviated month name, so shouldn't this
work?
print(date("M", mktime(date("M")-$i));
and if I loop;
Aug
Jul
Jun
May
........
[/snip]
Apparently not. But you can subtract by total hours in a given period, so
for an average I chose 30 days * 24 hours (720);
$i = 36;
$h = 720;
while($i > 0){
print(date("M", mktime(date("M")-$h)));print(date("Y"));print("<br>");
$i--;
$h = $h + 720;
}
which results in
Jul2002
Jun2002
May2002
Apr2002
Mar2002
Feb2002
Jan2002
Dec2002
Nov2002
Oct2002
Sep2002
Aug2002
Jul2002
Jun2002
May2002
So now all I have to do is sove the year rollover, which I think I can do
the same way. It'd be so much easier if mktime worked as the documentation
said, but after many iterations of different things this is what I was able
to come up with.
Jay
Carpe Ductum - Sieze the tape
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