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PHP Month Calucation

Posted on 2010-08-18
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Last Modified: 2013-12-13
I have data in date format mmm-yy like Apr-09. What I want to do is find out how old the date is from current date. I want to find out the number of months from current month.
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Question by:PranjalShah
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11 Comments
 
LVL 8

Expert Comment

by:ropenner
ID: 33469899
$input_date = "Apr-09";
$date = DateTime::createFromFormat('M-Y', $input_date);
$date_formatted = $date->format('Y-m-d');
$after = new DateTime(date("Y-m-d"));
$before = new DateTime($date_formatted);
$interval = $after->diff($before);
echo $interval->format('%y years + %m months + %d days');
0
 
LVL 5

Expert Comment

by:gyoreg
ID: 33469900
If you have >=5.3, then you can do this, assuming $olddatestring is filled with a date in the specified format:
$oldDate = DateTime::createFromFormat('M-y', $olddatestring, '0:0');

$diff = $oldDate->diff(new DateTime());

$diff_days = $diff->y * 12 + $diff->m;

// check diff docs at http://hu.php.net/manual/en/datetime.diff.php for further info...

Open in new window

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LVL 111

Expert Comment

by:Ray Paseur
ID: 33469906
Where did this date format come from?  Any reason you are not using the ISO8601 standard for the date format?
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LVL 8

Expert Comment

by:ropenner
ID: 33469918
Had an error above ... Y was supposed to be 'y' since your year is only two digits.

$input_date = "Apr-09";
$date = DateTime::createFromFormat('M-y', $input_date);
$date_formatted = $date->format('Y-m-d');
$after = new DateTime(date("Y-m-d"));
$before = new DateTime($date_formatted);
$interval = $after->diff($before);
echo $interval->format('%y years + %m months + %d days');


# output is:   1 years + 4 months + 0 days
0
 
LVL 8

Author Comment

by:PranjalShah
ID: 33469978
@Ray, I get the data from 3rd party so i dont have any control on what format they use.
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LVL 8

Author Comment

by:PranjalShah
ID: 33470017
I have PHP 5.2.10
0
 
LVL 17

Expert Comment

by:Shinesh Premrajan
ID: 33472183
$datetime1= DateTime::createFromFormat('M-Y', $input_date);
$datetime2= new DateTime();
$interval = $datetime1->diff($datetime2);

echo $interval->format('%R%d days');

Hope this helps
0
 
LVL 8

Author Comment

by:PranjalShah
ID: 33477711
I think the function createFromFormat is not supported in my version of PHP 5.2.10 ...Let me know if there is turn around
0
 
LVL 111

Accepted Solution

by:
Ray Paseur earned 2000 total points
ID: 33483710
Here is a function that will compute it correctly for you.  Months are a little tricky because (unlike weeks, days, hours, minutes) they do not have the same number of seconds.  But the PHP functions strtotime() and date() play together well enough to enable us to get the right answers by asking something like "Today - 1 month"

best regards, ~Ray
<?php // RAY_temp_months.php
error_reporting(E_ALL);
echo "<pre>" . PHP_EOL;

// A FUNCTION TO COUNT MONTHS
function months_ago($old)
{
    // MUNG THE DATA INTO A USABLE STRING AND GET TIMESTAMP
    $old = str_replace('-', ' 1, ', $old);
    $old_ts = strtotime($old);
    if ($old_ts === FALSE) return FALSE;

    // FIRST DAY OF THIS MONTH
    $new = date('M') . ' 1, ' . date('Y');
    $new_ts = strtotime($new);

    // MONTH COUNTER
    $months = 0;
    while ($new_ts > $old_ts)
    {
        $new_ts = strtotime(date('Y-m-d', $new_ts) . ' - 1 month');
        $months++;
    }

    return $months;
}

// TEST DATA FORMAT M-y AS SHOWN ON EE
// SEE http://us.php.net/manual/en/function.date.php
echo PHP_EOL . months_ago('Apr-09');

// SOME OTHER SAMPLES
echo PHP_EOL . months_ago('Aug-10');
echo PHP_EOL . months_ago('Jul-10');
echo PHP_EOL . months_ago('Jun-10');
echo PHP_EOL . months_ago('DEC-10');
echo PHP_EOL . months_ago('Jan-10');
echo PHP_EOL . months_ago('Sep-01');

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LVL 8

Author Closing Comment

by:PranjalShah
ID: 33489059
Thats perfect Ray, thanks
0
 
LVL 111

Expert Comment

by:Ray Paseur
ID: 33491409
Thanks for the points - it's a great question, ~Ray
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