Adding days to date, outputting incorrect dates?

Asked Jul 18 '18 at 13:05

Active Jul 18 '18 at 13:05

Viewed 21 times

I'm having an issue getting my code to add a day depending on the result of a drop down, the drop down values are 1-6 and the held in a variable $changeDate and depending on each adds the corresponding amount of days onto my current date. However, for some reason the output day isn't working correctly.

$currentDate = "01/02/2003";

if($changeDate == "1"){
    $newDate = date('d/m/Y', strtotime($currentDate.' + 1 days'));
}
elseif($changeDate == "2"){
    $newDate = date('d/m/Y', strtotime($currentDate.' + 2 days'));
}
elseif($changeDate == "3"){
    $newDate = date('d/m/Y', strtotime($currentDate.' + 3 days'));
}
elseif($changeDate == "4"){
    $newDate = date('d/m/Y', strtotime($currentDate.' + 4 days'));
}
elseif($changeDate == "5"){
    $newDate = date('d/m/Y', strtotime($currentDate.' + 5 days'));
}
elseif($changeDate == "6"){
    $newDate = date('d/m/Y', strtotime($currentDate.' + 6 days'));
}

echo $newDate;

Results of $changeDate == 1: 03/01/2003

Results should be: 02/02/2003

Results of $changeDate == 2: 04/01/2003

Results should be: 03/02/2003

Results of $changeDate == 3: 05/01/2003

Results should be: 04/02/2003

Results of $changeDate == 4: 06/01/2003

Results should be: 05/02/2003

Results of $changeDate == 5: 07/01/2003

Results should be: 06/02/2003

Results of $changeDate == 6: 08/01/2003

Results should be: 07/02/2003

asked Jul 18 '18 at 13:05

ben topper

Dates in nn/nn/nnnn format are considered MM/DD/YYYY. – John Conde Jul 18 '18 at 13:07
Dates with a `/` in are assumed to be the American style dates i.e. Month/Day/Year – RiggsFolly Jul 18 '18 at 13:07
If you convert `/` to `-` they will work as expected – RiggsFolly Jul 18 '18 at 13:09
So the best option would be to reformat `$currentDate` and after reformat `$newdate` – ben topper Jul 18 '18 at 13:10
Should just need to reformat `$currentdate` – RiggsFolly Jul 18 '18 at 13:11
change from $currentDate = "01/02/2003"; to $currentDate = "2003-02-01"; – ashish kumar Jul 18 '18 at 13:14
1

@RiggsFolly Perfect issue resolved! Thank you! – ben topper Jul 18 '18 at 13:17

Adding days to date, outputting incorrect dates?

0 Answers0