CODEWITHSUNDEEP
phpdatetimeadditiondateinterval
dft99
dft99

Reputation: 69

PHP Fatal error when trying to add a DateInterval to a DateTime object

What am I doing wrong here?

PHP Fatal error: Call to a member function add() on a non-object in

    $rndM = rand(48,51);
    $rndS = rand(1,59);
    $ts = new DateTime($krow['GLSTRT']);
    $tsd = $ts->format('Y-m-d H:i:s');
    $tsup = 'PT'.$rndM.'i'.$rndS.'s';
    $lamDate= $tsd->add(new DateInterval('\''.$tsup.'\''));

$krow['GLSTRT'] is a data value from MSSQL.

At first I was getting format errors with the add(new DateInterval).

I added in the format line, which resolved the format error and now the add(new DateInterval) is throwing the error at the beginning of this post.

Upvotes: 1

Views: 857

Answers (3)

dft99
dft99

Reputation: 69

:banghead: Thank you guys. This is what I get for rushing a fix. Here's the final product.

    $ts = new DateTime($krow['GLSTRT']);
    $intervalString = 'PT' . $rndM . 'M' . $rndS . 'S';
    $ts ->add(new DateInterval($intervalString));
    $lamDate = $ts->format('Y-m-d H:i:s');

Upvotes: 0

Jens A. Koch
Jens A. Koch

Reputation: 41756

  • The DateTime->format() has the following method signature:

    public string format ( string $format ). It returns a string.

  • You don't need to reassign the DataTime object to work with it.

I've modified the order of the code and the variable names a little bit to make things clear:

// create new datetime object and return formatted date
$date = new DateTime($krow['GLSTRT']);
echo $date->format('Y-m-d H:i:s') . "\n";

// define interval string and create a new DateInterval object
$rndM = rand(48,51);
$rndS = rand(1,59);
$intervalString = 'PT' . $rndM . 'i' . $rndS . 's';
$intervalObject = new DateInterval($intervalString); 

// add interval to date object  
$date ->add($intervalObject); 

// return the modified value formatted  
echo $date->format('Y-m-d H:i:s') . "\n";

Upvotes: 1

jeroen
jeroen

Reputation: 91734

A few things:

  • format() returns a string, so you cannot use add() on the result as it is not an object. You need $ts->add() instead.
  • You should check the right syntax for the DateInterval constructor; at the very least you need to get rid of the quotes around your string.
  • When you use the add() method, you modify your object, so assigning it to another variable is pointless unless you clone it first. Now both $ts (see my first point) and $lamDate will reference the same object.

Upvotes: 2

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