每次都创建新的datetime对象

I have created different vars using date_format from one datetime object $date_obj.

$date_obj = date_create($date);
$year = date_format($date_obj, 'Y');
$month = date_format($date_obj, 'm');
$day = date_format($date_obj, 'd');

However I have read (lost source) that this is bad practice? Instead I should create a new datetime object each time because the original object isn't referenced, but directly manipulated.

$date_obj_1 = date_create($date);
$year = date_format($date_obj_1, 'Y');

$date_obj_2 = date_create($date);
$month = date_format($date_obj_2, 'm');

$date_obj_3 = date_create($date);
$day = date_format($date_obj_3, 'd');

Is this true?

The DateTime object is an object, and therefore is passed by reference.

In your example above this won't matter, because you only format a date, you do not manipulate it. However, if you use a DateTime object as an argument in a function and inside this function you manipulate the object, your changes will be visible outside of that function:

function addDays($date,$days){
        $date->add(new DateInterval($days.' days'));
}

$date_obj_1 = date_create($date);

$formatedDate1 = date_format($date_obj_1, 'Y-m-d');
addDays($date_obj_1,10);
$formatedDate2 = date_format($date_obj_1, 'Y-m-d');

In the above example $formatedDate1 is different from $formatedDate1 because $date_obj_1 was passed by reference

EDIT: for a detailed explanation on my above snipped look at the comments section. @Xatoo explained it pretty good.

date_format isn't manipulating the DateTime object. What you do is the equivalent of:

$dateObject = new DateTime($date);
$year = $dateObject->format('Y');
$month = $dateObject->format('m');
$day = $dateObject->format('d');

This is absolutely fine, dateObject is not changed by calling the format method on it.