Add value to a date with PHP

Question

I’m having a problem adding up a date that comes from the database that comes in the following way: Ex.: 10/05/2018 17:48:27 I am using the following method to add days to that date:

date('d/m/Y H:i:s', strtotime('+10 days', strtotime('10/05/2018 17:48:27')));

But the result of what I get is this: 10-01-1970 16:00:00

I found that when the date format is with "-" (ie... 10-05-2018 17:48:27) instead of "/" the sum works.

My question is, how would make this sum work with the standard I receive from the database, as in the example "10/05/2018 17:48:27".

Answer 1

It’s always safer in these cases to use DateTime::createFromFormat, since it is you who tells PHP what format should be used to parse the date.

Behold:

 $datetime = DateTime::createFromFormat('d/m/Y H:i:s', '10/05/2018 17:48:27');

To add the days, just use modify

 $datetime->modify('+10 days');

To display use format

 echo $datetime->format('d/m/Y H:i:s');

Some comments on what you put in the question

But the result of what I get is this: 10-01-1970 16:00:00

Usually, when PHP cannot process a date correctly, the date is set. In some cases, the class DateTime even has errors in the date syntax (in this case, I believe the parser used internally from strtotime make that check);

If you see that example where I use the function date_parse, you can see how PHP understood its date string.

Sample Code:

 print_r(date_parse ('10/05/2018 17:48:27'));

The exit is:

array(12) {
  ["year"]=>
  int(2018)
  ["month"]=>
  int(10)
  ["day"]=>
  int(5)
  ["hour"]=>
  int(17)
  ["minute"]=>
  int(48)
  ["second"]=>
  int(27)
  ["fraction"]=>
  float(0)
  ["warning_count"]=>
  int(0)
  ["warnings"]=>
  array(0) {
  }
  ["error_count"]=>
  int(0)
  ["errors"]=>
  array(0) {
  }
  ["is_localtime"]=>
  bool(false)
}

Note that PHP interprets 05 as dia and the 10 as a month.

I found that when the date format is with "-" (ie... 10-05-2018 17:48:27) instead of "/" the sum works.

Yes, functions such as strtotime and date_parse or the class DateTime of PHP have a list of predefined formats to make the date interpretation, in which the format that is usually used in Brazil is not part of this list.

As I said before, just use createFromFormat that your problem is solved.