Php: Add variable amount days to Y-m-d date format

Question

i'm trying to add days to a date with the 'Y-m-d' format:

$oldDate = '2013-05-15';
$newDate = date('Y-m-d', strtotime($oldDate. " + 5 days"));

This ouputs '2013-5-20', but below:

$oldDate = '2013-05-15';
$addedDays = 5;
$newDate = date('Y-m-d', strtotime($oldDate. " + $addedDays days"));

doesn't work, it only outputs '1970-01-01', which doesn't make sense because i only tried to put the days to be added in a variable. They're basically the same code. I appreciate the help trying to understand this. Thanks!

Works for me: http://codepad.viper-7.com/pWNbA8 – John Conde Mar 17 '14 at 02:05 — John Conde, Mar 17 '14 at 02:05

score 8 · Accepted Answer · answered Mar 17 '14 at 02:08

8

However the code is right and works, just in case try

$newDate = date('Y-m-d', strtotime($oldDate. " + {$addedDays} days"));

answered Mar 17 '14 at 02:08

Javad

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Ooops, my mistake really. i was adding '-2' days that's why.. anyway thanks for verifying the code – muffin Mar 17 '14 at 02:13

score 0 · Answer 2 · answered Mar 17 '14 at 02:12

Use the DateTime class. It will spare you a lot of head-ache.

// Create a DateTime object
$date = new DateTime('2013-05-15');

// Original date
echo $date->format('d. F Y'), '<br>';

// Add 5 days
$date->modify('+5 days');

// Modified date
echo $date->format('d. F Y'), '<br>';

score -1 · Answer 3 · edited Mar 30 '15 at 17:42

-1

I had checked it, but it doesn't work on my computer (likely due to the PHP version). I found an alternative solution, though:

$timeBase = time();

$sDays2change = '+182'; // 6 months

$newtime = strtotime($sDays2change . ' day', $timeBase);

echo date('d/m/Y', $newtime);

edited Mar 30 '15 at 17:42

Chris Forrence

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answered Mar 30 '15 at 10:28

Colito

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Php: Add variable amount days to Y-m-d date format

3 Answers3