Add days to date variable

Question

I am trying to add days and/or weeks to a date that is being pulled from a database. All I am getting is the 12-31-1969 default date when it cannot output correctly. Here is my code:

$lastFeed = "6-25-2013"; //pulled from database last feed date 

$feedSchedule = "2"; //pulled from database number of weeks.

$nextFeeding = date("m-d-Y", strtotime($lastFeed . ' + ' . $feedSchedule. ' week'));

I have also tried multiplying the days times the $feedSchedule variable and replacing week(s) with day(s).

Sorry that was a typo I have edited it to replace it with $lastFeed. — n_starnes, Jun 25 '13 at 18:01
http://stackoverflow.com/questions/2332681/add-number-of-days-to-a-date — cristi _b, Jun 25 '13 at 18:04
I am trying to add days/weeks to a specific pulled time, not today's date which your link suggest. — n_starnes, Jun 25 '13 at 18:08
have you echo/print your retrieved `$lastFeed` variable to ensure that your year is not being retrieved as 0000 which would return 12-31-1969 — amaster, Jun 25 '13 at 18:17
I have. I just tried to format it using to see if it would output the same date("m-d-Y", strtotime($lastFeed)); and I'm getting the 1969 date. — n_starnes, Jun 25 '13 at 18:23
echo just the retrieved MySQL value, I bet that you are getting something similiar to `00-00-0000` — amaster, Jun 25 '13 at 18:29
I am not I have echoed the MySQL value and it is indeed showing the 6-25-2013 value. — n_starnes, Jun 25 '13 at 18:31

score 2 · Answer 1 · answered Jun 25 '13 at 18:03

2

6-25-2013 is not a valid date format. Try YYYY-MM-DD

answered Jun 25 '13 at 18:03

php_nub_qq

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The date format I am using works fine and for the purpose it is serving. I have used it in the past and on previous lines of code with this application. – n_starnes Jun 25 '13 at 18:06
@n_starnes But as you have observed, it doesn't work as an input format for `strtotime()`. You don't have to switch the format globally, but you will need to modify it for use with `strtotime()`. – George Cummins Jun 25 '13 at 18:07
doing it as you suggest displays: 1969-Dec-Wed I need accurate date information not day of the week. – n_starnes Jun 25 '13 at 18:15

amaster · Accepted Answer · 2013-06-25T18:41:15.430

0

Here is code that will work and account for the invalid Date Time String

function nextFeeding($lastFeed,$feedSchedule){
  //fix date format
  $correctedDate = explode("-",$lastFeed);
  //pad month to two digits may need to do this with day also
  if($correctedDate[0] < 10 && strlen($correctedDate[0])!==2){
    $correctedDate[0] = "0".$correctedDate[0];
  }
  $correctedDate = $correctedDate[2]."-".$correctedDate[0]."-".$correctedDate[1];
  //get the next feeding date
  $nextFeeding = date("m-d-Y", strtotime($correctedDate . ' + ' . $feedSchedule. ' week'));
  //return value
  return $nextFeeding;
}

$lastFeed = "6-25-2013"; //pulled from database last feed date 

$feedSchedule = "2"; //pulled from database number of weeks.

$nextFeeding = nextFeeding($lastFeed,$feedSchedule);
echo $nextFeeding;

returns

07-09-2013

edited Jun 25 '13 at 18:41

answered Jun 25 '13 at 18:15

amaster

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Sorry, I forgot to put the semicolon on the third line... just tried it in PHP sandbox and all works! – amaster Jun 25 '13 at 18:21
I placed the semicolon there because I noticed it as well. It still does not work. – n_starnes Jun 25 '13 at 18:23
try this before running the conversion `echo $lastFeed;` and let me know exactly what you are getting. – amaster Jun 25 '13 at 18:25
I am running this without a conversion. $lastFeed when echoed shows the correct format of 6-25-2013 and then $lastFeed is then used in your function. Once the function is run it shows the date as 12-31-1969. $feedschedule varies each time it is used. On one animal it is 2 on the other it can be 10. – n_starnes Jun 25 '13 at 18:30
I did put in the date and the raw number instead of a variable and it returned it just fine. – n_starnes Jun 25 '13 at 18:33
can you edit your question to include all of your code including mysql query... please remember security though and do not include your database credentials... I believe you may have a typo with a variable somewhere if it works when you define the variable hard coded vs defining it with your mysql query – amaster Jun 25 '13 at 18:36
I echoed the $correctedDate after the conversion in the function and it is showing a formated 6-25-2013 from the variable as 2013-006-25. Any ideas? – n_starnes Jun 25 '13 at 18:36
I have verified that I am pulling the MySQL data correctly and echoed/printed it onto the webpage successfully. It is going through your function but is not adding the weeks to the date successfully. – n_starnes Jun 25 '13 at 18:38
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/32354/discussion-between-amaster507-and-n-starnes) – amaster Jun 25 '13 at 18:41

Add days to date variable

2 Answers2