Date time convert is incorrectly being modified

Question

I am having issues getting my code to return the correct response.

$Birthd = '06-27-1996';

$NewISSdate = date("m-d-Y", strtotime(date("m-d-Y", strtotime($Birthd)) . " +21 years"));

When I run this the response is: "12-31-1969"

I believe this is a default date of sorts, but what can I do to repair my code? If ran with a different $Birthd string such as "07-03-1996".

And 0 is 1970/01/01 so 31st of December is 0 epoch with timezone correction - also look here: https://stackoverflow.com/questions/18269556/php-date-add-1-year-to-current-date — mplungjan, Jul 21 '17 at 05:37

score 0 · Answer 1 · answered Jul 21 '17 at 05:36

0

Change - to / and try...

$Birthd = '06/27/1996';

$NewISSdate = date("m/d/Y", strtotime(date("m/d/Y", strtotime($Birthd)) . " +21 years"));

answered Jul 21 '17 at 05:36

Arshad Shaikh

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try with this `date("m/d/Y",strtotime('+21 years', strtotime($Birthd)));` – Arshad Shaikh Jul 21 '17 at 05:45

Alive to die - Anant · Accepted Answer · 2017-07-21T07:22:28.123

0

You need to change string date format and then add years to it like below:-

$Birthd = '06-27-1996';
$Birthd = str_replace("-","/",$Birthd);

echo $NewISSdate = date("Y-m-d", strtotime(date("Y-m-d", strtotime($Birthd)) . " + 21 years"));

Output:-https://eval.in/835509 OR https://eval.in/835555

Reference:- php different acceptable date formats

edited Jul 21 '17 at 07:22

answered Jul 21 '17 at 05:37

Alive to die - Anant

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@AlivetoDie I think Im in love. You saved my life. Thank you! – Ryan Reddings Jul 21 '17 at 05:54
@RyanReddings glad to help you :):) – Alive to die - Anant Jul 21 '17 at 05:55

Date time convert is incorrectly being modified

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