I need to calculate at which year a school class has started, given the year they currently are in. For example a class of 2nd graders have started school in 2013, if I pass 2 to the function it should return 2013.

So, I need to take x years and 6 months out of a date, I was wondering which is the better way to do it.

Using strtotime():

function school_started1($years){
    return date('Y', strtotime('now - '.$years.' year - 6 month'));

Or calculating it manually:

function school_started2($years){
    $seconds_in_half_a_year = 365.25 * 43200;
    return date('Y', time() - $years * 365.25 * 86400 - $seconds_in_half_a_year);

I'm thinking the second method would be better because it doesn't have to do any parsing and stuff ( not quite a fan of parsing ). I can do the math myself, are there any benefits/downsides to using strtotime? What do you think?


1 Answer 1


The first version, using strtotime(), is definitely more maintainable. The code obviously says exactly what you intend. I like it.

The second version is full of magic numbers (even though they are somewhat recognizable magic numbers), and is therefore not as clear. I don't like it, but if you wanted to use it anyway, I suggest:

function school_started2_alt($years){
    $SECONDS_PER_YEAR = 365.25 * 86400;
    return date('Y', time() - ($years + 0.5) * $SECONDS_PER_YEAR);

If you are using PHP ≥ 5.3, also consider this formulation, which flows nicely from left to right:

(new DateTime())->sub(new DateInterval("P${years}Y6M"))->format('Y');

The kinds of strings accepted by the DateInterval constructor follow a rigid format, so the parsing should be efficient. Keep in mind that PHP itself is an interpreted language, so the work to parse a DateInterval specifier string is insignificant relative to the total amount of parsing involved. You should therefore take full advantage of PHP's built-in features to simplify your own code.


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