Solved

php year month loop function

Posted on 2012-12-30
3
2,060 Views
Last Modified: 2012-12-30
$curYear = date('Y');
          for ($k = $curYear; $k >= 2011; $k--)

          for ($i = 12; $i >= 1; $i--)

$monthName = date("F", mktime(0, 0, 0, $i, 10));

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what i want to modify in this code is

like if year = 2012 i will start from 2012 and the current month ...and decrement the month till January..



and if year less than curr year like 2011 fetch all month..


let say we are in year 2013 so the output is
2013 from month 12 to 1
and 2012 from month 12 to 1
and 2011 from month 12 to 1

i want to modify it to be like
2013 from current  month  to 1
and 2012 from month 12 to 1
and 2011 from month 12 to 1
0
Comment
Question by:afifosh
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3 Comments
 
LVL 13

Expert Comment

by:darren-w-
ID: 38730967
well to get months this year to today:

<?php 
$thisYear = range(1,date("n"));
foreach ($thisYear as $month){
print date("F",mktime(0,0,0,$month,10));
}

//previous years :

$thisYear = range(1,12); //then just add years

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0
 
LVL 110

Expert Comment

by:Ray Paseur
ID: 38730986
Please read the article here.  It will tell you everything you need to know about the basics of handling PHP and SQL date/time values.
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_201-Handling-date-and-time-in-PHP-and-MySQL.html
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LVL 110

Accepted Solution

by:
Ray Paseur earned 500 total points
ID: 38730996
Here is how I might do it.
<?php // RAY_temp_afifosh.php
error_reporting(E_ALL);
ini_set('display_errors', TRUE);
date_default_timezone_set('America/Chicago');
echo '<pre>';

/* PROBLEM DEFINITION
i want to modify it to be like
2013 from current  month  to 1
and 2012 from month 12 to 1
and 2011 from month 12 to 1
*/

// A DATE IN 2013
$start = '2013-04-15';

// A DATE TO STOP GENERATING DATA
$stop  = '2011-01-01';

// A LOOP TO SHOW THE DATES
while ($start >= $stop)
{
    $start = date('Y-m-01', strtotime($start));
    echo PHP_EOL . $start;
    $start = date('Y-m-01', strtotime($start . ' - 1 MONTH'));
}

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Best to all, ~Ray
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