[RESOLVED] Factoring current grade based on grad year

jasonmills58

Hey all,

So, I figured there must be a way to do this, but my brain is fried. To cut down on maintenance, I'm tracking 500 or so students via their graduation year, rather than their current grade, so I don't have to go in every summer and ratchet up all the grades. Not that hard to do, I know, but I like good dynamic systems.

So anyway, I need to allow the teachers to search for them based on grade (9, 10, 11, or 12). So what I need is to be able to compare the grade chosen on any given day of the year, and have the code figure out what grad year that person is in, assuming a constant of the 4 year plan, and do a SELECT * WHERE gradYear = 2009.

The problem is that I could take the nowYear or something, but that changes half-way through the school year (stupid northern hemisphere) and I'm stumped on how to compare more complicated dates. Making a cut-off date of June 30th would work, but I've totally confused myself.

Any help?
Thanks much,
Jason

harmor

//what year is it now?
$year = date("Y");

//what grade is the student in?
$grade = 9;

//substract their current grade by "12"
$grade = "12 - $grade";

//we add the new value of "$grade" to the current "$year"
$year = "$year + $grade";

/*
If the above code worked as I suspect I should get "2009" as the graduation year.
*/


$getstudent = mysql_query("SELECT * FROM students WHERE gradYear = '".intval($year)."' ");
$stu = mysql_fetch_array($getstudent);

Installer

As I read the specs, anyone who is a ninth-grader today (after June 30) will graduate in 2010.

This will do it:

$grade = '9';

$today_doy = date('z');
$june_30_doy = date('z', strtotime(date('Y') . '-06-30'));
if ($today_doy <= $june_30_doy) {
    $grad_year = date('Y') + 12 - $grade;
} else {
    $grad_year = date('Y') + 13 - $grade;
}

$query = "SELECT * WHERE gradYear = '" . $grad_year . "'";

jasonmills58

Thanks Installer - I'll give that a shot.

Thank you also Harmor, but the problem you'll run into with that is when the $year = date("Y"); changes in January.

But all helpful ideas - my brain thanks you!

I'm open to other suggestions as well.

harmor

Good algorithm Installer.

Good luck, Jason

jasonmills58

Hey Installer,

Correct me if I'm wrong, but since $grade in your example never changes, couldn't we just change

if ($today_doy <= $june_30_doy) {
    $grad_year = date('Y') + 12 - $grade;
} else {
    $grad_year = date('Y') + 13 - $grade;
}

if ($today_doy <= $june_30_doy) {
    $grad_year = date('Y') + 3;
} else {
    $grad_year = date('Y') + 4;
}

Just a thought - thanks again!

EDIT - Nevermind, I get it - "9" was passed from the selector as chosen by the teacher!

harmor

No, it will not work.
What if the grade is "10" or higher?

Edit: Just saw your edit.

jasonmills58

Thanks much both of you - this oughta do it.

Installer

harmor -

What if the grade is "10" or higher?

It'll work for any grade 0 through 12. 🙂

harmor

Installer wrote:
harmor -It'll work for any grade 0 through 12. 🙂

The reason why I said "10 or higher" is because he only wants grades 9-12.

Installer

You've lost me. Good night.