Auto increment years in loop every year [duplicate] - php

Question

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Closed 10 years ago.
Possible Duplicate:
How to increment date with 1 (day/year) in PHP?
Im not really sure where to begin with this but im trying to make a year start at 1928 and stop at 1948 and for every year the years increment by one so since its 2012 the date ranges are 1928 - 1948 and for 2013 it would be 1929 - 1949 and 2014 would be 1930 - 1950 and so on...
right now i just have a basic loop for when to start and stop the years but its not too dynamic, like i said im pretty much at a blank on where to begin other then date('Y')+1.
for($i=1928;$i<=date('Y');$i++)
{
echo '<option value='.$i.'>'.$i.'</option>';
if($i == '1948'){break;}
}

Answer 1

So you want it to go between current year minus 84 and the current year minus 64? Use this code:
$firstYear = (int)date('Y') - 84;
$lastYear = $firstYear + 20;
for($i=$firstYear;$i<=$lastYear;$i++)
{
echo '<option value='.$i.'>'.$i.'</option>';
}
Edit: updated for performance. Current year is determined before the loop (per Pitchinnate's comment).

Answer 2

Try this:
$year = date('Y');
$add = $year - 2012;
$min = 1928 + $add;
$max = $min + 20;
for($i=$min;$i<=$max;$i++)
{
echo '<option value='.$i.'>'.$i.'</option>';
}
I isn't a good idea to have date('Y') or any evaluations done on the for loop as it gets calculated every time through the loop. Article about this.

Answer 3

Something like this?
$base_year = 2012;
$start_year = $base_year - 84;
$end_year = $start_year + 20;
for( $i = $start_year; $i <= $end_year; $i++)
{
echo '<option value='.$i.'>'.$i.'</option>';
}

Answer 4

for($i =0; $i <= 20 ;$i++)
{
$year = date('Y') - 84 + $i;
echo '<option value='.$year.'>'.$year.'</option>';
}

Answer 5

It looks like you have a constant with how far back you want the date range to be from the current year that you can use to write a loop that will work for you.
Since it's 2012 now, and you want the range to start at 1928 when it's 2012, then we can use 2012 - 1928 = 84, so the year that starts the range should always be 84 less than the current year.
Therefore we could write code like:
$startingYear = date('Y') - 84;
$endingYear = $startingYear + 20;
for ($i = $startingYear;$i <= $endingYear;$i++)
{
echo '<option value='.$i.'>'.$i.'</option>';
}