calculation problem, someone good in math?

hi there,

im on a image gallery here, and i wanna display x new thumbs per day. if i dont have any new images to show, i wanna go back to the first picture and start over again.
so if i have 10 images and wanna show 4 new each day, i must display:
day1: 1-4
day2: 5-8
day3: 1-4 (forget 9 and 10, just start over again)
day4: 5-8
...
every thumb has a database entry, and for this problem i have this values:
$thumbrate - number of thumbs to show each day, 4 in this case
$age - age of this gallery in days
$counter - total number or thumbs in this category, 10 here

i need: $offset, the number of results to skip, must be 0 for day1 and day3, 4 on day 2 and day 4.

can someone think about a formula to solve this problem please, or contact me on icq? thanks...

 

// how many have we already shown?
if ( $age ) {
    $alreadyShown = $age*$thumbrate;
    
    // show next or start over?
    if ( $alreadyShown+$thumbrate < $counter ) {
        $offset = $alreadyShown;
    } else {
        $offset = 0;
    }
} else {
    $offset = 0;
}

PHP:

I think that works... assuming $age starts at 0 rather than 1.

 

hm ok, you solved the easy part - the problem is after "else": that keeps it forever on $offset=0 after the first circle - but it should repeat the whole gallery over and over again.

 

You're right, my fault. Working version:

// if age is 0, offset 0, otherwise...
if ( $age ) {
	// days for a cycle
	$daysPerCycle = floor($counter/$thumbrate);
	// days occured so far in current cycle
	$inCurrent = $age % $daysPerCycle;
	// images already shown in this cycle
	$alreadyShown = $inCurrent*$thumbrate;

	// show next or start over?
	if ( $alreadyShown+$thumbrate < $counter ) {
		$offset = $alreadyShown;
	} else {
		$offset = 0;
	}
} else {
    $offset = 0;
}

PHP:

Hopefully that's better.

 

excellent - its even very close the my non-working solution